1 straight line-1 12-07-2015 - cbsecontents s.no. topics page number 1. coordinate geometry - i 1 2...

PREFACE

Welcome to UDAAN - a program to give wings to girl students!

The UDAAN programme has been initiated with the primary objective of increasing

the enrollment of girl students in leading engineering institutions. The program is

designed to provide a comprehensive platform to deserving girls who aspire to pursue

higher education in engineering and assist them in preparing for the entrance

examinations and increasing their probability of being a student in the most prestigious

engineering institutions of the country.

This study material provided to you is aligned with the tutorials given to you on your

tablets and made available online. A soft copy of this material is available on your

tablets as well. You will find that the material has been simplified and made easy to

read. Each topic has been divided into Subtopics to relate it with the tutorials. Each

subtopic discussed is followed by some practice questions for you to attempt. Answers

to these questions are given at the end. In case you have difficulty in solving any of

them, please contact your helpdesk and seek clarifications.

You can call 1800 180 7542 or write to [email protected] for any queries.

Best of luck and happy learning !

Chairperson, CBSE

Contents

S.No. Topics Page Number

1. Coordinate Geometry - I 1

2 Coordinate Geometry - II 5

Practice Questions 9

3. Coordinate Geometry - III 11

4. Coordinate Geometry - IV 19

Practice Question 20

5. Coordinate Geometry - V 23


6. Coordinate Geometry - VI 27

7. Coordinate Geometry - VII 33

8. Coordinate Geometry - VIII 37


9. Circle: Equation of Circles - I 45

10. Circle: Position of a Point - II 49

11. Circle: Problem Solving - III 51


12. Circle: Tangents and Normals - IV 57

13. Circle: Tangents and Normals - V 61

14. Circle: Tangents and Normals - VI 65


15. Circle: Intersection of Two Circles - VII 79

16. Circle: Radical Axis - VIII 85

17. Circle: -Co- Axial System of Circles - IX 91

18. Circle: Limiting Points- X 95

19. Circle: Problem Solving - IX 99


20. Parabola Conic Section - I 111


Contents

S.No. Topics Page Number

21. Parabola Parabola - II 117


22. Parabola -III 121


23. Parabola Parabola - IV 125


24. Parabola Parabola - V 129

25. Parabola Parabola - VI 135


26. Parabola Parabola - VII 141

27. Parabola Parabola - VIII 143

28. Parabola Parabola - IX 147


29. Parabola Parabola - X 153


1

COORDINATE GEOMETRY-IStraight Line

Topics covered1. Slope of a straight line2. Angle between two lines3. Special forms of straight lines

(i) Slope and point form(ii)Two point form(iii) Slope and y-intercept form

Straight Line -It is a curve such that every point on the line segment joining any two points on it lies on it.Equation of straight line is always of first degree on x and y.General equation of straight line is ax+by+c=0Where a,b,c R.a and b can be zero but not both at the same time .

Slope of a lineThe trigonometrical tangent of an angle that a line makes with the positive direction of the x-axis in anticlockwise direction is called the slope or gradient of the lineNormally slope is denoted by m= tan θ .

x

y

O

θ

y

x

θO

Points to remember(i) Slope of a line parallel to x-axis is zero and perpendicular to x-a xis is undefined.(ii) If slope is positive then it makes acute angle with positive direction of x-axis and nega-

tive it will make obtuse angle with positive direction of x-axis.(iii) If a line is equally inclined to the axes, that it will make angle of 450 or 1350, with x-

axis. Hence slope of line is + 1.

(iv) It three points A,B,C are collinear then slope of AB=slope of BC=slope of AC

(v) Slope of the line joining two points (x1, y

1) and (x

2, y

2) is m=

12

12

21

21

xx

yyor

xx

yy

−−

−−

x

θ

B

C M

y

A N

O

(x , y )1 1

(x , y )2 2

(vi) Equation of x-axis is y=0(vii) Equation of y-axis is x=0

∈

2

y=a (a>0)

y=a (a<0)

x

y

O

(viii) Equation of line parallel to x-axis is y = a, where a is any real number(ix) Equation of line parallel to y-axis (perpendicular to x-axis) is x=a, where a is any real

number

x=a(a>0)x=a(a<0)

O

y

x

Angle between two linesThe angle θ between the lines having slopes m

1 and m

2 is given by

tan θ = ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+−+=

+−

21

21

21

21

mm1

mm

mm1

mm

O

y

x1

B A

θ1θ2

θP

Where m1=tan θ 1

, m2 =tan θ 2

(i) When two lines are parallel then their slopes are equal ie. m1= m

2

(ii) When two lines are perpendicnlar then the product of their slope is -1i.e. m

1m

2=-1 or 1+m

1m

2=0

A line equally inclined with two linesLet the two lines with slopes m

1 and m

2 be equally inclined to a line with slope m then

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+−

−=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+−

mm1

mm

mm1

mm

2

2

1

1

m1

m1

mθ

θ

Special form of a line1. Point slope form

The equation of a line which passes through the point (x1,y

1) and has the slope m is y-y

1=m(x-

x1)

3

o

y

x

A(x , y )1 1

P(x, y)

2. Two point formThe equation of a line passing through two points (x

1,y

1) and (x

2,y

2) is

y-y1= ( )1

12

12 xxxx

yy−

−−

o

y

x

A(x , y )1 1

B(x , y )2 2

P(x, y)

3. Slope and y intercept form [Non Vertical Line]The equation of a line with slope m that makes an intercept c on y-axis is y=mx+c

Oθ

y

x

(o, c)

Example

1. Find the equation of a line that has y-intercept -3 and is perpendicular to the line joining (2,-3) and (4,2).

Solution

Slope of line joining the points (2,–3) and (4,2) is 2

5

24

32

xx

yy

12

12 =−+=

−−

Since the required line is perpendicular to it

∴ (slope of the required line) ( )25X�

= -1

∴ slope of the required line = 52− =m

y-intercept = -3

Equation of line is y = 3x52 −−

2x+5y+15 = 0

4

Examples:1. If the centroid and circumcentre of a triangle are (3,3) and (6,2) respectively, then the orthocentre

is(a) (-3,5) (b) (-3,1) (c) (3,-1) (d) (9,5)Solution: Centroid, circumcentre and orthocenter are collinear such that centroid divides thecircumcentre and orthocentre in the ratio 1:2. Ans(-3,5).

2. If the algebraic sum of the perpendicular distances from the point (2,0), (0,2) and (1,1) to avariable straight line be zero, then the line passes through the point(a) (3,3) (b) (1,1) (c) (1,-1) (d) (-1,-1)Solution: Let equation be ax+by+c=0

A.T.Q. 0ba

cba

ba

cb20

ba

coba2222222

=+

++++

++++

++

⇒3a+3b+3c=0 ⇒ a+b+c=0This shows that ax+by+c=0 passes through (1,1).

3. The area enclosed by 2|x|+3|y|≤6 is(a) 3 Sq-units (b) 12 Sq.units (c) 9 Sq.units (d) 24Sq.unitsSolution: 2x+3y ≤ 6, x≥0, y≥02x-3y≤ 6, x≥0, y≤0-2x+3y ≤ 6, x≤0, y≥0-2x-3y ≤ 6, x≤0, y≤0Form a rhombus with diagonals 4 & 6.

Area =2

1×4×6= 12 sq. units.

1

2

3

4

5

COORDINATE GEOMETRY-II

Straight Line

Topics Covered1 Intercept form2 Normal form3 Distance form OR Symmetric form OR Parametric form

Intercept formThe equation of a straight line which cuts off intercepts a and b on x–axis and y–axis respec-tively is given by

y

(0, b)

O (a, 0)x

y

(0, b)

O(–a, 0)

x

y

(0, –b)

O(–a, 0)x

y

(0, –b)

O (a, 0)x

b

y

a

x + = 1b

y

a

x– + = 1b

y

a

x + = –1b

y–

a

x = 1

Normal form (perpendicular form)The equation of a straight line upon which the length of the perpendicular from the origin is p andthe perpendicular makes an angle α with the positive direction of x–axis is given byxcos α +ysin α = p

y

B

A

pα

Ox

Note Here p is always taken as positive and α is measured from positive direction of x–axis inanticlockwise direction between 0 and 2 π

Distance form or Symmetric form or Parametric FormThe equation of a straight line passing through the point (x

1, y

1 ) and making an angle θ with the

positive direction of x–axis is given by

θ=

θ sin

y–y

cos

x–x 11 = r

Where r is the distance of the point (x, y) from the point (x1,y

1)

Parametric formFrom the above equation we getx–x

1 = rcos θ , y–y

1 = rsinθ

x = x1+rcos θ y=y

1+rsinθ

y

B

r

A(x ,y )1 1

P (x,y)

O

θ

θ x

6

The coordinates (x,y) of any point on the line at a distance r from the point A (x1,y

1) can be taken as

(x1+rcosθ , y

1+rsinθ )

Where the line is inclined at an angle θ with positive direction of x–axis.

1 If P is on the right side of A(x1,y

1) then r is positive and if p is on the left side of A(x

1,y

1) then

r is negative.

2 At a given distance r from the point (x1,y

1) on the line θ

=θ sin

y–y

cos

x–x 11 , there are two points

viz; (x1+rcosθ ,y

1+rsinθ ) and (x

1–rcosθ ,y

1–rsinθ )

Examples

1 Through the point P( α 1β ) where α β >0 the straight line b

y

a

x + = 1 is drawn so as to form with

coordinate axes a triangle of area S. If ab>0 then the least value of S is

(a) α β (b) 2 α β (c) 4 α β (d) None.Solution

Given equation of line is b

y

a

x + = 1 ____________________(1)

area of Δ OAB = S

ab2

1 = S

Or ab = 2S ab>0

Equation (1) passes through the point P(α ,β )

ba

β+α = 1 or

ab

a

a

β+α = 1

S2

a

a

β+α = 1

a2β –2aS+2S α = 0a is real

∴D≥0

4S2–8S α β ≥ 0

S2–2S α β ≥ 0

S≥ 2 α β

∴ least value of S is 2 α β .2 The distance of the point (2,3) from the line 3x+2y =17 measured parallel to the line x-y=4 is

(a) 24 (b) 25 (c) 2 (d) None.

Solution.Any point of the line parallel to x–y = 4 and passes through P(2,3) at a distance r, are

y

B(0, b)

OA(a, 0)

b

a

SP( , )α β

x

7

(2+rcos θ ,3+rsin θ ) where tan θ = 1=slope of line x–y =4 therefore θ = 45° = 4π

This point lies on the line 3x+2y = 17∴ 6+3r cos45° + 6 + 2r sin45° = 17

6+2

r3+ 6 +

2

r2 = 17

2

r5 = 5

2

r = 1

r = 2

3 A ray of light travelling along the line x+ 3 y = 5 is incident on the x-axis and after refraction it

enters the other side of the x–axis by turning π /6 away from the x–axis. The equation of the linealong which the refracted ray travels is

(a) x+ 3 y–5 3 = 0 (b) x– 3 y–5 3 = 0

(c) 3 x+y–5 3 =0 (d) 3 x–y–5 3 =0

Solution : The refracted ray passes through the point (5, 0) and makes an angle 1200 with positivedirection of x–axis∴ The equation of the refracted ray isy–0 = tan1200(x–5)

y = – 3 (x–5)

3 x+y–5 3 = 0

4 Column Matching (This Q. is not relevant as option the righthand coloumn are unique)easily identifide.

Column I Column II1 It a, b, c are in H.P, then the straight line

c

1

b

y

a

x ++ = 0 always passes through a (a)2

π

fixed point2 The larger of the two angles made with

the axis of a straight line drawn through (b) (1, –2)(1, 2) so that it intersects x+y = 4 at a

point distant3

6 from (1, 2) is

3 The diagonals of the parallelogramwhose sides are l x+my+n = 0,lx+my+n1 = 0, mx+ly+n = 0 (c) Rhombus

x–y=4

3x+2y=17

P(2, 3)

y

Retracted Ray

(5. 0)

b

Retracted Ray

.16

8

mx+ly+n1 = 0 include an angle4 The diagonals of a parallelogram

PQRS are along the lines x+3y = 4and

6x–2y = 7. Then PQRS must be a (d)125π

Solution :

1 a, b, c are in H.P so a

1,

b

1,

c

1 are in A.P

∴ b

2 =

a

1 +

c

1

or a

1 –

b

2 +

c

1 = 0 =

a

x +

b

y +

c

1

on comparing x = 1, y = –2 so coordinate (1,–2)

2 AP = r = 3

6

Equation of AP ⇒ θcos

1–x = θsin

2–y = r

∴P (1+r cos θ , 2+rsinθ )P will satisfy x+y = 4

∴1+3

6 cos θ + 2+

3

6 sin θ = 4

cos( θ – 4

π) =

2

3 = cos30°

∴ θ = 75° or 15° ie, 12

5π or

12

π

3 Since the distance between parallel lines Lx+my+n=0 and Lx+my+n’=0 is same as the

distance between the parallel lines mx+ly+n = 0 and mx+ly+n1 = 0. ie, 22

–

mL

nn

+

′ there

fore the parallelogram is a rhombus. Since the diagonals of a rhombus are at right angles,

therefore the required angle is 2

π.

4. Slope of line x+3y = 4 is 3

1–

slope of line 6x–2y = 7 is 2

6 = 3

product of slopes = 3

1– × 3 = –1

S

P P

Rx+3y=4

6x–2y=7

y

A(1,2)

x + y=4

Prθ

Ox

9

∴ diagonals are perpendicular to each otherAnswers

1 ⇔ b 2 ↔ d3 ↔ a 4 ⇔ c

PRACTICE QUESTIONS

1 Orthocenter of triangle with vertices (0, 0), (3, 4) and (4, 0 ) is

(a) ⎟

⎠

⎞

⎜

⎝

⎛

4

5,3 (b) (3, 12) (c) ⎟

⎠

⎞

⎜

⎝

⎛

4

3,3 (d) (3, 9)

2 Area of the triangle formed by the line x+y = 3 and angle bisectors of the pairs of straight linesx2–y2 + 2y = 1 is(a) 2 sq.units (b) 4 sq.units (c) 6sq.units (d) 8 sq.units

3 Let O (0, 0), P(3, 4), Q (6, 0) be the vertices of the triangle OPQ. The point R inside the Δ OPQis such that the triangle OPR, PQR, OQR are of equal area. The coordinates of R are

(a) ⎟

⎠

⎞

⎜

⎝

⎛ 3,3

4(b) ⎟

⎠

⎞

⎜

⎝

⎛

3

2,3 (c) ⎟

⎠

⎞

⎜

⎝

⎛

3

4,3 (d) ⎟

⎠

⎞

⎜

⎝

⎛

3

2,

3

4

4 Consider three points P(–sin(β –α ),–cosβ ), Q(cos(β –α ), sinβ ) and R(cos(β – α +θ ), sin(β –

θ )), where 0< α , β , θ <4

π. Then

(a) P lies on the line segment RQ.(b) Q lies on the line segment PR.(c) R lies on the line segment QP.(d) P, Q, R are non-collinear.

5 The locus of the orthocenter of the triangle formed by the lines (1+p) x– py + p (1+p) = 0, (1+q)x– qy + (1+q)q = 0 and y = 0, where p ≠ q, is(a) a hyperbola (b) a parabola (c) an ellipse (d) a straight line.

6 Let points A (1, 1) and B (2, 3). Coordinates of the point P such that |PA–PB| is minimum are

(a) ⎟

⎠

⎞

⎜

⎝

⎛

2

3,2 (b) ⎟

⎠

⎞

⎜

⎝

⎛

4

11,0 (c) (11, 3) (d) ⎟

⎠

⎞

⎜

⎝

⎛ 0,2

3

7 The line x+7y = 14 is rotated through an angle 4

π in the anticlock wise direction about the point (0,

2). The equation of the line in its new position is(a) 3x–4y+8 = 0 (b) 3x–4y–8 = 0 (c) 4x+3y+8 = 0 (d) None of these

8 If one diagonal of a square is the portion of the line b

y

a

x + = 1 intercepted by the axes, then one

to the extremities of the other diagonal of the square are

(A) ⎟

⎠

⎞

⎜

⎝

⎛ +2

b–a,

2

ba(b) ⎟

⎠

⎞

⎜

⎝

⎛ +2

ba,

2

b–a (c) ⎟

⎠

⎞

⎜

⎝

⎛

2

a–b,

2

b–a (d) ⎟

⎠

⎞

⎜

⎝

⎛ +2

a–b,

2

ba

10

9 The image of P(a, b) in the line y = –x is Q and the image of Q in the line y=x is R, then the midpointof PR is

(a) (a–b, b+a) (b) ⎟

⎠

⎞

⎜

⎝

⎛ ++2

ab,

2

ba(c) (a–b, b–a) (d) (0, 0)

10 Let ABC be a triangle. Let A be the point (1, 2), y = x is the perpendicular bisector of AB and x–2y+1 = 0 is the angle bisector of ∠ C. If equation of BC is given by ax+by – 5 = 0, then a+b is(a) 1 (b) 2 (c) 3 (d) 4

ANSWERS1 c 2 a 3 c 4 d 5 d 6 b7 a 8 c 9 d 10 b

11

COORDINATE GEOMETRY-IIIStraight Line

Topics Covered1 Distance of a line from a point2 Distance between two parallel lines3 Area of a parallelogram4 Area of triangle

1 Distance of a line from a pointLength of perpendicular (distance) from the point (x

1, y

1) to the line ax+by+c = 0 is given by

p = 22

11

ba

|cbyax|

+++

y

B

O

HA

P(x ,y )1 1

x

0,

a

c–

bc

,–0

Length of perpendicular from the origin to the line ax+by+c = 0 is given by

p = 22 ba

|c|

+ , ( x1, y

1) is (0, 0)

y

O

b

a

pμ

x

0,

a

c–

b

c,–0

2 Distance between two parallel linesThe distance between two parallel lines ax+by+c

1 = 0 and ax+by+c

2 = 0 is given by

p = 22

21

ba

|c–c|

+

ax+by+c =01

ax+by+c =02

12

Note that coefficient of x and y of both equation must be same.3 Area of a parallelogram

Area of a parallelogram ABCD = 2 area of Δ ABD

A=

If the sides of a parallelogram be a1x+b

1y+c

1 = 0 ; a

1x+b

1y+d

1 = 0 :a

2x+b

2y+c

2 = 0, a

2x+b

2y+d

2 =

0 area of parallelogram =

22

11

2211 )–)(–(

ba

badcdc

Area of Rhombus

In case of a rhombus p1 = p

2, area of rhombus =

θsin

21p

A

D

B

C

d1d2

A

D

B

C

P1

P2

Also area of rhombus = 2

1d

1d

2 where d

1, d

2 are the lengths of two perpendicular diagonals

Area of a triangleArea of a triangle whose vertices are (x

1, y

1), (x

2, y

2) and (x

3, y

3) is

2

1 |x

1(y

2–y

3) + x

2(y

3–y

1) + x

3(y

2–y

1)|

sin θ sin θ

P

sin θ

P 12xA ⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛=

θ

D

B

C

p1

p2

Aθ

⎟

⎠

⎞

⎜

⎝

⎛= sinθ x AD x ABx 2

1 zA

sinθ AD

P2 and sinθ

AD

PL ===Here

A= θsin21 pp

sinθ x AD x AB A =

13

Or 2

1

11

33

22

11

yx

yx

yx

yx

= 2

1[(x

1y

2+x

2y

3+x

3y

1) – (x

2y

1+x

3y

2+x

1y

3)|

1 If area of a triangle is given then use ± sign.2 If three points A, B, C are collinear then area of triangle ABC is zero.3 If a

1x+ b

1y+c

1 = 0 ; a

2x+b

2y+c

2 = 0 and a

3x+b

3y+c

3 = 0 are the three sides of a triangle, then

the area of the triangle is given by

Δ =

2

333

222

111

321 cba

cba

cba

CCC2

1

where C1 = a

2b

3–a

3b

2

C2 = a

3b

1–a

1b

3

C3 = a

1b

1 –a

2b

1

C1,C

2,C

3 are cofactors of c

1, c

2 & c

3

Area of polygonThe area of polygon whose vertices are (x

1, y

1), (x

2y

2) .............(x

n1y

n)

= 2

1

11

nn

22

11

yx

yx

.

.

.

yx

yx =

2

1[(x

1y

2+x

2y

3.............x

ny

1) – (y

1x

2+y

2x

3...........+y

nx

1)]

Example1 If the area of the rhombus by the lines x ± my ± n = 0 be 2sq. units then

(a) , m, n are in GP (b) , n, m are in GP(c) m = n (d) L = m.

Solution : (b)By solving the sides of the rhombus, we get the vertices of rhombus are (0, –n/m), (–n/ ,o),(0,n/m)

and (n/ ,0). Hence the area is 21

×mn2

× n2

= 2

∴ n2 = m.

14

Hence , n, m are in GP.2 The area of the triangular region in the first quadrant bounded on the left by the y –axis;bounded above by the line 7x+4y = 168 and bounded below by the line 5x+3y = 121 is A, then the

value of 10

A3 is

y

x

P(20,7)

(0,42)

20

0

3

121,0

0,7

168

0,5

121

SolutionGiven lines 7x+4y = 168 and 5x+3y = 121 intersect at P (20,7) by solving these equationsArea of shaded region is

A =2

1⎟

⎠

⎞

⎜

⎝

⎛

31

40–42 20

=2

1 ×

3

5× 20 =

3

50 sq.units

10

3A =

10

3×

3

50 = 5

3 Column I Column IIa Four lines x+3y–10 = 0 p a quadrilateral which is neither

x+3y–20 = 0, 3x–y+5 = 0 and parallelogram nor a trapezium3x–y–5 = 0 form a figure which is

b The points A(1, 2), B(2,– 3), C(–1,–5) q a parallelogramand D(–2, 4) in order are the vertices of

c The lines 7x+3y–33 = 0, 3x–7y+19 = 0, r a rectangle of area 10sq.units3x–7y–10 = 0 and 7x+3y–4 = 0 form afigure which is

d Four lines 4y–3x–7 = 0,3y–4x+7 = 0, 4y–3x–21 = 0,3y–4x+14 = 0 form a figure s a squarewhich is

Solutiona →q, r, s b →p c →q,s d →q

a h1 =

10

10 = 10

h2 =

10

10 = 10

x+3y–10=0

x+3y=20

h1h2

3x–y+5=03x–y–5=0

15

Hence the given lines form a square of side 10 . Area of a square is ( )210 = 10sq.units

b slop of AB = 1–

5 = – 5

slop of DC = 1–

9 = – 9

slop of AD = 3

2– =

3

2–

slop of BC = 3

2

Hence the figure is neither a parallelogram nor a trapezium

c h1 =

58

29 =

2

29

h2 =

58

29 =

2

29

Hence given lines form a square of side 2

29 and area

229

sq.units

d 4y–3x–21 = 0 and 4y–3x–7 = 0 are parallel and 3y–4x+14 = 0 and 3y–4x+7 = 0 are parallel.But 4y–3x–21 = 0 and 3y–4x+14 = 0 are not perpendicularHcnce the given line form a parallelogram.

4 The straight lines 7x–2y+10 =0 and 7x+2y–10 = 0 form an isosceles triangle with the line y = 2.Area of this triangle is equal to

(a)7

15 sq.units (b)

7

10 sq.units

(c)7

18 sq.units (d) None

Solution

B ⎟

⎠

⎞

⎜

⎝

⎛ 2,7

6 and C ⎟

⎠

⎞

⎜

⎝

⎛ 2,7

6–

7x+3y–4=0

7x+3y–33=0

h1h2

3x–7y+19=03x–7y–10=0

A (0,5)

y=2C BD

y

x

2,

7

6–

2,7

6

0,7

10

0,

7

10–

B(2,–3)B(2,–3)

A(1,2)

D(–2,4)

C(–1,–5)

O

4y 3x 21=0

4y 3x 7=0

3y 4x+7=03y 4x+14=0

16

BC = 7

12 and AD = 3

area of Δ ABC =2

1×

7

12 × 3

=7

18sq.units.

Linked comprehension typeFor problems 5–7Let ABCD be a parallelogram whose equations for the diagonals AC and BD are x+2y = 3 and2x+y = 3, respectively. If length of diagonal AC = 4 units and area of parallelogram ABCD = 8sq.units. then

5 The length of other diagonal BD is(a) 103 (b) 2 (c) 20/3 (d) None

6 The length of side AB is equal to

(a)3

582(b)

9

584(c)

9

583(d)

9

585

7 The length of BC is equal to

(a)3

102(B)

3

104(C)

3

108(d) None

Solution :

5 tan θ =

11

221–

+

+ = 4

3

sin θ = 53

, cosθ = 5

4

Area of Δ CPB = 2

1 PC × PB sin θ

4

1 × 8 =

2

1 × 2 × PB ×

5

3

PB = 3

10

BD = 3

20 units

A

x+2y=3

2x+y=3

D

B

C

P θπ θ–

17

6 cos( π–θ ) = PBAP2

AB–PBAP 222

×+

–cos θ = 5

4− =

3

1022

AB–9

1004 2

××

+

AB = 3

582

7 In Δ CPB, cos θ = PBPC

BCPBPC

..2

– 222 +

BC = 3

102

8 ReasoningType

Statement 1 : The area of the triangle formed by the points A(1000, 1002), B(1001, 1004) C(1002,1003) is same as the area formed by A1(0, 0), B1(1, 2), C1(2, 1)

Statement 2 : The area of the triangle is constant with respect to translation of axes.(a) both the statements are true but statement 2 is the correct explanation of statement 1(b) both the statement are true but statement 2 is not the correct explanation of statement1(c) statement 1 is true and 2 is false(d) statement 1 is false and 2 is true

Solution :

Area of triangles is unaltered by shifting origin to any point. If origin is shifted to (1000, 1002) A, B,C become P(0, 0), Q(1, 2), R(2, 1) both are true and statement 2 is correct -explanation of state-ment 1.Ans. a

19

COORDINATE GEOMETRY-IVStraight Line

Topics Covered1. General Equation of straight line2. Reducing general equations to

i. Slope intercept formii. Intercept formiii. Normal form

General Equation of straight lineFirst degree equation of the formax+by+c=0 where a,b,c R a and b can be zerobut not both at the same time.

Reducing general equation to slope intercept formGiven equation is ax+by+c=0Rewrite the equation by =-ax – c

divide by b we get b

c–x

b

a–

It look like y = mx+c

slope =b

a−= – m

yoftCoefficien

xoftCoefficien =

y- intereept =b

c−= –

y oft Coefficien

Constant

Reducing to intercept formGiven equation is ax+by+c=0

Rewrite the equation cbyax �−=+

1b/c

y

a/c

x =−

+−

1b

y

a

x =+

So intercepts are -c/a & -c/b on x-axis and y-axis respectively.

Reduce to normal formGiven equation be ax + by + c = 0Re write the equation ax + by = – c

– ax – by = cKeeping constant term postive

Divide by 22 ba + we get 222222––

ba

cy

ba

bx

ba

a

+=

++

∈

20

It look like pαsiny αcos x =+

Where ( )a

bαtanor

ba

b–αsin,

ba

a–αcos

2222=

+=

+=

distance of a line from the origin is always positive

22 ba

c

+∴ = p is positive.

PRACTICE QUESTIONS

1. A triangle ABC with vertices A(–1, 0), B ⎟

⎠

⎞

⎜

⎝

⎛

4

3,2– & C ⎟

⎠

⎞

⎜

⎝

⎛

6

7,–3– has its orthocenter H. Then the

orthocentre of triangle BCH will be(a) (–1, 0) (b) (–3, –2)(c) (1,3) (d) (–1, 2)

2. The points A(0, 0), B(cos α , sinα ) and C(cosβ , sinβ ) are the vertices of a right angled triangleif

(a) sin ⎟

⎠

⎞

⎜

⎝

⎛ β+α2

= 2

1(b) cos ⎟

⎠

⎞

⎜

⎝

⎛ β+α2

= 2

1(c) cos ⎟

⎠

⎞

⎜

⎝

⎛ βα2

– =

2

1

(d) sin ⎟

⎠

⎞

⎜

⎝

⎛ β+α2

= – 2

1

3. Set of values of α for which the point (α , α 2–2) lies inside the triangle formed by the lines x+y =1, y = x+1 and y = –1 is

(a) ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ +⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

2

131–,1U1,–

2

13–1(b) ( )13,1

(c) ( )1,–13– (d) None

4. Area of the parallelogram formed by the lines y = mx, y = mx+1, y = nx+1and y= nx equals

(a) 2)n–m(

|nm| +(b) |nm|

2

+ (c) |nm|

1

+ (d) |n–m|

1

5. A and B are fixed points. The vertex C of Δ ABC moves such that cotA+cotB = constant. Thelocus of C is a straight line(a) perpendicular to AB (b) parallel to AB (c) inclined at an angle (A–B) to AB(d) None of these.

6. The area of the figure formed by a|x|+b|y| +c = 0 is

(a)|ab|

c2

(b)|ab|

c2 2

(c)|ab|2

c2

(d) None of these

7. The orthocentre, circumcentre, centroid and incentre of the triangle formed by the line x+y = a with

21

the coordinate axes lie on(a) x2+y2 = 1 (b) y = x (c) y = 2x (d) y = 3x

8. Two points (a, 3) and (5, b) are the opposite vertices of a rectangle. If the other two vertices lie onthe line y = 2x + c which passes through the point (a, b) then the value of c is(a) –7 (b) –4 (c) 0 (d) 7.

ANSWERS1. a 2. c 3. a 4. d 5. b 6. b7. b 8. a

23

COORDINATE GEOMETRY-VStraight Line

Topics Covered1. Image of a point in different cases.2. Foot of the perpendicular.3. Family of lines.

Image of a point in different casesi. The image of a point with respect to the line mirror

Image of A(x1,

y1) with respect to line mirror ax+by+c=0 be B(x

2, y

2) is given by

22

11212

ba

)cbyax(2

b

yy

a

xx 1

+++

=−=−

M is the foot of prependicular from A on ax+by+c = 0

B(x ,y )2 2

A(x ,y )11

M

ax+by+c=0

ii. The image of a point with respect to x-axis:-Let A(x

1, y

1) be any point and B(x

2, y

2) its image after reflection in the x-axis then.

x1 = x

2 & y

2= –y

1 (M is the mid point of A&B)

y

M

A(x ,y )11

B(x ,y )22

Ox

M is the foot of perpendicular from A on x-axisiii. The image of a point with respect to y-axis:-

Let A (x1,

y1) be any point and B (x

2, y

2) its image after reflection in the mirror y-axis then.

x2= –x

1 & y

1=y

2 (N is the mid point of A&B)

N is the foot of perpendicular from A on y-axis

y

N

A(x ,y )11B(x ,y )22

Ox

24

iv. The image of a point with respect to the origin:-Let A(x

1, y

1) be any point B (x

2, y

2) be its image after reflection through the origin

then.x

2= –x

1 & y

2= –y

1 (O is the mid point of A&B)

y

N M

A(x ,y )11

B(x ,y )22

O x

v. The image of a point with respect to the line y = xLet A(x

1, y


2, y

2) be its image after reflection in the line y=x then.

y2=x

1 and x

2=y

1 (M is the mid point of A&B)

y

M

A(x ,y )11

y=x

B(x ,y )22

45°O x

vi The image of a point with respect to the line y = x tan θLet A (x

1, y


2, y

2) be its image after reflection in the line

y=x tan θ or y=mx then.x

2=x

1 cos2 θ +y

1 sin2 θ (M is the mid point of A&B)

y2=x

1 cos2 θ –y

1 sin2 θ

y

M

A(x ,y )11

y=xtanθ

B(x ,y )22

θO x

Line parallel and perpendicular to given lineGiven equation of straight line be ax+by+c=0A line parallel to given line is ax+by+d=0 only constant term changes.A line perpendicular to given line is bx-ay+k=0Here change coordinate of x as coordinate of y & coordinate of y as negative of coordinate

of x and constant term changesIt the lines a

1x+b

1y+c

1 = 0 and a

2x+b

2y+c

2 = 0 are prependecular then a

1a

2+b

1b

2= 0

i. Coincident, if 2

1

2

1

2

1

c

c

b

b

a

a ==

25

ii. Parallel, if 2

1

2

1

2

1

c

c

b

b

a

a ≠=

iii. Intersecting, if 2

1

2

1

b

b

a

a ≠

PRACTICE QUESTIONS

1. If t1 and t

2 are roots of the equation t2+λ t+1 = 0, where λ is an arbitrary constant. Then the

line joining the points (at12, 2at

1) & (at2

2, 2at

2) always passes through a fixed point

(a) (a, 0) (b) (–a, 0) (c) (0, a) (d) (0,–a)2. The equation x3+y3 = 0 represents

(a) three real straight lines (b) three points (c) combined equation of a st. line & acircle (d) None of these.

3. The three lines whose combined equation is y3 – 4x2y = 0 form a triangle which is(a) isosceles (b) equilateral (c) right angled (d) None of these

Comprehension Type

A(1, 3) and C ⎟

⎠

⎞

⎜

⎝

⎛

5

2,–

5

2– are the vertices of a triangle ABC and the equation of the angle bisector

of ∠ ABC is x+y = 24 Equation of side BC is

(a) 7x+3y = 4 (b) 7x+3y+4 = 0 (c) 7x–3y+ 4 = 0 (d) 7x–3y = 45. Coordinates of vertex B are

(a) ⎟

⎠

⎞

⎜

⎝

⎛

10

17,

10

3(b) ⎟

⎠

⎞

⎜

⎝

⎛

10

3,

10

17(c) ⎟

⎠

⎞

⎜

⎝

⎛

2

9,

2

5– (d) (1, 1)

6. Equation of side AB is(a) 3x+7y=24 (b) 3x+7y+24 = 0 (c) 13x+7y+8 = 0 (d) 13x–7y+8 = 0

7. Assertion and reasoning TypeLines L

1 L

2 given by y –x = 0 and 2x + y = 0 intersect the line L

3 given by y+2 = 0 at P and Q,

respectively. The bisector of the acute angle between L1 and L

2 intersects L

3 at R.

Statement 1 : The ratio PR : RQ equals 22 : 5 .

Statement 2 : In any triangle, bisector of an angle divides the triangle into two similar triangles.a Statement 1 is true, Statement 2 is True; Statement 2 is a correct explanation for State-

ment1.b Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for

Statement 1.c Statement 1 is True, Statement 2 is False.d Statement 1 is False, Statement 2 is True.

8. Matrix-matchThis question contains statements given in two columns which have to be matched. Statements a,b, c, d in column I have to be matched with statements p,q, r, s in column II. If the correct matchis a →p, a →s, b →q b →r, c →p,c →q and d →s, then the correctly dubbled 4×4 matrix

26

should be as follows :Consider the lines given by

L1 : x + 3y – 5 = 0

L2 : 3x – ky –1 = 0

L3 : 5x + 2y –12 = 0

Column I Column IIa L

1, L

2, L

3 are concurrent, if p. k = – 9

b One of L1, L

2, L

3 is parallel to at least one of

the other two, if q. k = – 6/5c L

1, L

2, L

3 form a triangle, if r. k = 5/6

d L1, L

2, L

3 do not form a triangle, if s. k = 5

ANSWERS1 b 2 d 3 d 4 b 5 c 6 a

7 c 8 q,pb

sa

→→

s,q,pd

rc

→→

27

COORDINATE GEOMETRY-VIStraight Line

Topics Covered1. Concurrent lines.2. Equations of angles bisectors.3. Position of two points relative to a given line.4. Standard points of triangle.5. Equations of straight lines through (x

1, y

1) making angle α with line y=mx+c

1. Concurrent linesThe three given lines are concurrent if they meet at one point.i. Find the point of intersection of any two lines by solving them simultaneously. If this

point satisfies the third equation also then the given lines are concurrent.ii. Let three lines be

L1=a

1x+b

1y+c

1= 0 L

2=a

2x+b

2y+c

2=0 and

L3=a

3x+b

3y+c

3= 0 are concurrent if

0

cba

cba

cba

333

222

111

=��

��

��

iii. The three lines L1=0, L

2=0 and L

3=0 are concurrent if there exist constants , m and n

not all zero at the same time, such that

L1+mL

2+nL

3= 0

2. Eqnations of angle bisectors between two linesThe eqnations of the bisectors of the angles between the lines a

1x+b

1y+c

1= 0 and a

2x+b

2y+c

2=0

are given by 22

22

222

21

21

111

ba

cybxa

ba

cybxa

+++

±=+

++

i. Any point on a bisector is equidistant from the given lines.ii. Locus of points which are equidistant from the two intersecting lines is an angle bisector.iii. Bisectors are perpendicular to each other.iv. Equation of the bisector of the acute and of obtuse angle between two lines.

Let a2x+b

2y+c

2= 0 ......(1)

and a2+b

2y+c

2=0 ......(2)

c1>0,c

2>0 then the equation

22

22

222

21

21

111

ba

cybxa

ba

cybxa

++++=

+++

is the bisector of the acute or obtuse angle between the lines 1&2 according asa

1a

2+b

1b

2 < 0 or > 0

again 22

22

222

21

21

111

ba

cybxa–

ba

cybxa

+++=

+++

28

is the bisector of the acute or obtuse angle between the lines 1&2 according asa

1a

2+b

1b

2 > 0 or < 0

3. Position of two points relative to a given lineLet the line be L =ax+by+c=0P(x

1,y

1) and A(x

2,y

2) are two given points.

i. If ax1+by

1+c and ax

2+by

2+c both are of the same sign and hence cbyax

cbyax

++

++

22

11 >0 then

the points P and Q lie on same side of line ax+by+c=0

P

Q

L

P

Q

L

ii. If ax1+by

1+c and ax

2+by

2+c are of opposite sign and hence cbyax

cbyax

++

++

22

11<0 then the

points P and Q lie on opposite side of the line ax+by+c=0

P

Q

L

iii. If origin lie on line then the line is know as origin side.

P(x ,y )1 1

(0, 0)L

P(x ,y )1 1

O(0, 0)

L

iv. A point (x1y

1) will lie on origin side of the line ax+by+c=0 if ax

1+by

1+c and c have same

sign.v. A point (x

1, y

1) will lie on non-origin side of the line ax+by+c=0 if ax

1+by

1+c and c have

opposite sign

Family of straight linesLet L

1 = a

1x+b

1y+c

1=0 and L

2 =a

2x+b

2y+c

2=0

Then the general equation of any straight line passing through the point of intersection of lines L1

and L2

is given by L1+ λ L

2=0 where λ is any real number.

Equations of straight lines through (x1,y1) making angle α with y=mx+c

y-y1= )(

tan1

tan––)(

tan1

tan111 xx

m

myyandxx

m

m −+

=−−

+αα

αα

��

29

θα

α

θ α–

L2

L1

θ

Standard points of a triangle1. Centroid or centre of gravity :- The centroid of a triangle is the point of intersection of its

median’s. the centroid divides the median in the ratio 2:1 (vertex:base).

Coordinates of G are

++++

3

yyy,

3

xxxG 321321

A(x ,y )1 1

B(x ,y )2 2

F

1

2E

D

G

C(x ,y )3 3

In isosceles triangle median to the equal sides are equal in length and in equilateral triangle allmedians are equal in length.Equations of median can be obtained by using two point form with vertex and the mid point ofopposite side.

2. Circumcentre:- The circumcentre of a triangle is the point of intersection of the perpendicularbisectors of the sides of a triangle.Coordinates of O can be obtained from the equation.OA2 =OB2=OC2

A(x ,y )1 1

B(x ,y )2 2

O

C(x ,y )3 3

If angles A,B,C and vertices A(x1,y

1) , B(x

2,y

2) and C(x

3,y

3) of a ABCΔ are given; Then its

circumcentre is given by

++++

++++

C2sinB2sinA2sin

C2sinyB2sinyA2siny,

C2sinB2sinA2sin

C2sinxB2sinxA2sinx 321321

The circumcentre of a right-angled triangle is the mid-point of its hypotenuse. Therefore themid-point of hypotenuse is equidstant from its verticesAM=BM=CM

30

A

CB

H

The circumcentre of the triangle formed by (0,0), (x1,y

1) and (x

2,y

2) is

( ) ( )( )

( ) ( )( )

−

+−+−

+−+

2112

22

221

21

212

1221

22

221

21

21

yxyx2

yxxyxx,

yxyx2

yxyyxy 2

3. Incentre of a triangle:- The point of intersection of the internal bisecters of the angles of atriangle is called the incentre of the triangle.The coordinates of the incentre of a triangle with vertices (x

1, y

1), (x

2, y

2) and (x

3, y

3) are

++++

++++

cba

cybyay,

cba

cxbxax 321321

Where a,b,c are lengths of sides of triangle

A

a

bc

B

I

C

The incentre of the triangle formed by (0,0), (a,0), (0,b) is

++++++ 2baba

ab,

baba

ab222

4. Orthotcentre:- The orthocentre of a triangle is a point intersection of altitudes.i. Take the equations of any two sides of a triangle. find the eqnations the lines

perpendicular to those lines and passing through the opposite vertices. solve these twoequations we get orthocentre of the triangle.

ii. If angles A,B,C and vertices A (x1, y

1) B (x

2, y

2) and C (x

3, y

3) of a ABCΔ are given

then orthocentre of ABCΔ is given by

++++

++++

CtanBtanAtan

CtanyBtanyAtany,

CtanBtanAtan

CtanxBtanxAtanx 321321

iii. If any two lines out of three lines AB, BC, CA, are perpendicular, then orthocentre is thepoint of intersection of two perpendicular lines.

iv. The orthocentre of the triangle with verties (0,0), (x1, y

1) and (x

2, y

2) is

31

( ) ( )

−+−

−−−

1221

212121

2112

212121 yxyx

yyxxxx,

yxyx

yyxxyy ��

v. The orthocentre (O), centroid (G) and circumcentre (C) of any triangle lie in a straightline and G divides the join of O and C in the Ratio 2 :1

vi. In an equilateral triangle,orthocentre, centroid, circumcentre and incentre coinside.

5. Coordinates of nine point circle :- If a circle passes through the feet of perpendicular (D,E,F)mid points of sides BC, CA, AB sespectivcly (H,I,J) and mid points of the line joining theorthocentre O to the angular points A,B,C (K,L,M) thus the nine points D,E,F H,I,J, K,L,M alllie on a circle, This circle is known as nine point circle and its centre is called nine point centre.

A

B CD

E

FJ

K

I

ML

H

1. The orthocentre (O), nine point centre (N) centroid (G) and circumcentre (C) all lie in thesame line i.e. ONGC (oil natural gas corporation)

2. The nine point centre bisects the join of orthocentre (O) and circumcentre (C)3. The radius of nine point circle is half the radius of circumcircle.

33

COORDINATE GEOMETRY-VIIStraight Line

Examples1. Locus of the image of the point (2,3) in the line (x-2y+3)+λ(2x-3y+4)=0 is

(a) x2+y2-3x-4y-4=0(b) 2x2+2y2+2x+4y-7=0(c) x2+y2-2x-4y+4=0(d) None

Solution:

=+−=+−

04y3x2

03y2xP (1,2)

Let image be B(h,k) then AP=BP(h-1)2+(k-2)2 =12+12

h2+k2-2h-4k+5=2x2+y2-2x-4y+3=0

A(2,3)B

(h,k)

P

(1,2)

Ans. : d2. If the lines ax+y+1=0, x+by+1=0 and x+y+c=0 (a,b,c being distinct and different from 1) are

concurrent, then =−

+−

+⎟⎠

⎞

⎜

⎝

⎛

− c1

1

b1

1

a1

1

(a) 0 (b) 1 (c)cba

1

++ (d) None

Solution: c

2→c

2–c

1,c

3→c

3–c

1

0

1–c01

01–b1

a–1a–1a

0

c11

1b1

11a

=⇒=

a(b-1)(c-1)-(1-a)(c-1)+(1-a)(1-b)=0divide by (1-a)(1-b)(1-c) we get

⇒ 0c1

1

b1

1

a1

a =−

+−

+−

⇒ 1c1

1

b1

1

a1

1 =−

+−

+−

34

3. The set of values of ‘b’ for which the origin and the point (1,1) lie on the same side of the st. linea2x+aby+1=0 ∀ a ∈R, b>o are(a) b∈[2,4) (b) b∈(0,2) (c) b∈[0,2] (d) None of these

Solution: (0,0) & (1,1) are on the same side0+1>0 so a2+ab+1>0

⇒ D<0 i.e 4b

04b2

2

<<−

-2<b<2 but b>0∴ b∈(0,2)

4. Let P(-1,0), Q(0,0) and R ( )33,3 be three points. Then the equation of the bisector of the

angle PQR is

(a) 0yx2

3 =+ (b) 0y3x =+ (c) 0yx3 =+ (d) 0y2

3x =+

Solution:

01

m

m31

m3

+=

+−

( )( )3,

3

1m

0331m3

03mm3m3

03m2m3m3mm32

22

−=

=+−

=−−+

=−+⇒+=−

0yx3

x3y

=+

−=

R

m

P m=0 Q

3m =

5. OPQR is a square and M, N are the mid points of the sides PQ and QR respedively. If the ratio

of the areas of the square and the triangle OMN is λ : 6, then4

λ is equal to

(a) 2 (b) 4 (c) 12 (d) 16

35

Solution: ar. of square = a2

ar of Δ OMN = 2

18

a3

4

a3.

2

1

1a2

a

12

aa

10022

==

166

8

a3

a2

2

=λ⇒

λ=

Ans : b

R

O aX

Q

P

Y

a,2

aN

2

aa,M

6. A pair of perpendicular straight lines drawn through the origin form an inosceles triangle withline 2x+3y=6, then area of the triangle so formed is

(a)13

36(b)

17

12(c)

5

13(d)

13

17

Solution: OM=13

6

94

6 =+

−, PQ=

13

62 × , area od ΔOPQ = 13

36

13

6

13

62

2

1 =×××

o

PM

Q

(0,0)

2x+3y-6=0

Q

P

O

M45

4545

45

36

7. Match the columnColumn I Column II

(a) Two vertices of a triangle are (5,–1) and (–2,3) if orthocentre (p) (-4,-7) is origin then third vertex is

(b) A point on the line x+y=4 which lies at a unit distance from (q) (-7,11)the line 4x+3y=10, is

(c) Orthocentre of the triangle made by the (r) (1,-2)lines x+y-1=0, x-y+3=0, 2x+y=7 is

(d) If a,b,c are in A.P., then lines ax+by=c always Pass through (s) (-1,2)

Solution: A-pB-qC-sD-s

37

COORDINATE GEOMETRY-VIIIStraight Line

Examples

1. The number of integer values of m for which the x- coordinate of the point of intersection of the lines3x+4y=9 and y=mx+1 is also an integer is

(a) 2 (b) 0 (c) 4 (d) 1

Solution: Point of intersection ⎟

⎠

⎞

⎜

⎝

⎛

++

+ m43

m93,

m43

5

x-coordinate is an integer when 3+4m = ±1 or ±5.

m = -1,-2,2

1,

2

1−

Hence there are two values of m.2. A straight line through the origin meets the parallel lines 4x+2y=9 and 2x+y+6=0 at points P

& Q respectively. Then the point O divides the segment PQ in the ratio(a) 1:2 (b) 3:4 (c) 2:1 (d) 4:3Solution: Clearly Δ OPA~ Δ OQC.

⇒ 4

3

34

9

OC

OA

OQ

OP ===

3:4

o

D(0,-6)

C(-3,0)

Q

P

⎟

⎠

⎞

⎜

⎝

⎛

2

90,B

⎟

⎠

⎞

⎜

⎝

⎛0,

4

9A

3. A family of lines is given by (1+2λ)x+(1-λ)y+λ=0, λ being the parameter. The line belongingto this family at the maximum distance from the pt (1,4) is(a) 4x-y+1=0 (b) 33x+12y+7=0 (c) 12x+33y=7 (d) None of these.Solution: x+y+λ(2x-y+1)=0

The required line is y⇒⎟

⎠

⎞

⎜

⎝

⎛ ++

−=−

3

1x

3

11

3

14

3

1y

12x+33y=7.

P(1,4)

x+y=0

2x-y+1=0⎟

⎠

⎞

⎜

⎝

⎛−3

1,

3

1

38

4. The four sides of a quadrilateral are given by the equation xy(x–2)(y-3)=0. The equation of the lineparallel to x-4y=0 that divides the quadrilateral in two equal areas is(a) x-4y+5=0 (b) x-4y-5=0 (c) 4y=x+1 (d) 4y+1=x.

Solution: ar.of OADE=2

1ar OABC

2

1(OE+AD)×OA =

2

1×OA×AB

⎟

⎠

⎞

⎜

⎝

⎛ λ+λ−4

–2

42

1×2 =

2

1×2×3

4

22 λ− =3

λ∴ =-5∴Equation is x-4y+5=0

C

x-4y=0

B

DE

O A

x-4y=λ

5. Consider the points A(0,1) and B(2,0) and P be a point on the line 4x+3y+9=0. Coordinates ofP such that |PA–PB| is maximum are

(a) ⎟

⎠

⎞

⎜

⎝

⎛ −5

17,

5

24(b) ⎟

⎠

⎞

⎜

⎝

⎛ −5

13,

5

18(c) ⎟

⎠

⎞

⎜

⎝

⎛

7

31,

7

31(d) ( )0,0

Solution: |PA-PB|≤AB

Thus |PA-PB| is max. if points A,B,P are collinear. Equation of AB is y-1=12

10

−−

(x-0)⇒x+2y-

2=0

Hence solving x+2y-2=0 & 4x+3y+9=0 we get ⎟

⎠

⎞

⎜

⎝

⎛ −5

17,

5

24

39

PA(0, 1)

B(2, 0)

P1

4x+3y+9=0

6. A light ray coming along the line 3x+4y=5 gets reflected from the line ax+by=1 and goesalong the line 5x-12y=10. Then

(a)15

112b,

115

64a == (b)

115

8b,

15

14a

−== (c)115

8b,

115

64a

−==

(d)15

14b,

15

64a ==

Solution: ax+by=1 will be one of the bisectors of the lines given

⎟

⎠

⎞

⎜

⎝

⎛ −−+=−+13

10y12x5

5

5y4x3

⇒ 64x-8y=115 or 14x+112y=15 On comparing with ax+by=1, we get

115

8b,

115

64a

−== or15

112b,

15

14a ==

7. If the point (a,a) is placed in between the lines |x+y|=4, then a is(a) [-2,0] (b) [0,2] (c) (-2,2) (d) [-2,2]

Solution: x+y=4x+y=-4& pt (2,2) & (-2,-2) lies on these linesThen pt. (a,a) lies between the lines then a>-2 and a<2 ie. -2<a<2.

Ans: (C)

Ans: (C)

40

(2,2)

y=x

x+y=2

x+y=-2

(-2,-2)

8. Consider the family of lines 5x+3y-2+λ1(3x-y-4)=0 and x-y+1+λ2(2x-y-2)=0. The equationof a straight line that belongs to both the families is _______________________.Solution: Equation of a line belongs to both families

passes through A and B is

1y

1x

4yx3

2y3x5

−==

=−=+

4y

3x

2yx2

1yx

==

=−−=− ( )

( )07y2x5

1x2

51y

1x13

141y

=−−

−=+

−−+=+

A(1,–1)

B(3,4)

9. If x1, x

2, x

3 as well as y

1, y

2, y

3 are in G.P. with same common ratio then the points P(x

1,

y1),Q(x

2,y

2) and R (x

3,y

3)

(a) lie on a straight line(b) lie on an ellipse(c) lie on a circle(d) are vertices of a triangle.

Solution: Let x1=a, x

2=ar, x

3=ar2

y1=b, y

2=br, y

3=br2

41

Now a

b

xx

yy

12

12 =−−

& a

b

xx

yy

23

23 =−−

slope of PQ = slope of QRHence pts P,Q,R are collinear.Ans. : a

10. A variable straight line is drawn through the point of intersection of the straight lines 1b

y

a

x =+

and 1a

y

b

x =+ and meets the coordinate axes at A and B, find the locus of the midpoint of AB.

Solution: Let mid pt. of AB be (h,k).

Intersection pt. of given lines is ⎟

⎠

⎞

⎜

⎝

⎛

++ ba

ab,

ba

ab

P is mid pt. of AB ⇒ A(2h,0) & B(0,2k).

Now A,B and Q are collinear ⇒0

1ba

ab

ba

ab1k20

10h2

=

++

⇒ 4hk ba

hab2–

+ 0ba

kab2– =

+⇒ 2xy(a+b) = ab(x+y).

PRACTICE QUESTIONS

1. If the distance of any point (x,y) from origin is defined as d(x, y) =⏐x⏐+⏐y⏐ then the locus of d(x,y)=1 is aa. circle of area π sq. units b. square of area 1 sq. unitsc. square of area 2 sq. units d. none of the above

2. The line 3x–4y+7 = 0 is rotated through an angle 4

π in the clockwise direction about the point

(–1, 1). The equation of the line in its new position isa. 7y+x–6=0 b. 7y–x–6=0 c. 7y+x+6=0 d. 7y–x+6=0

3. The number of integral values of m for which the x-coordinate of the point of intersection ofthe lines 3x +4y = 9 and y = mx +1 is also an integer isa. 2 b. 0 c. 4 d. 1

4. One of the bisector of the angle between the lines a(x–1)2+2h(x–1)(y–1) + b(y–2)2=0 is x+2y–5=0 The other bisectora. 2x–y=0 b. 2x+y=0 c. 2x+y–4=0 d. x–2y+3=0

5. The point A(2, 1) is translated parallel to the line x–y=3 by a distance 4 units. If the newposition A is in third quadrant, then the coordinates of A’ are

P(h,k)

B

Q

(0, b)

(0, a)

(a, 0) (b, 0)

A

42

a. (2+2 2 , 1+2 2 ) b. (–2+ 2 , –1–2 2 )

c. (2–2 2 , 1–2 2 ) d. none

6. The orthocenter of the triangle formed by the lines x+y=1, 2x+3y=6 and4x–y+4=0 lies ina. I quadrant b. II quadrant c. III quadrant d. IV quadrant

7. The reflection of the point (4, –13) in the line 5x+y+6=0 isa. (–1, –14) b. (3, 4) c. (1, 2) d. (–4, 13)

8. If p1, p

2, p

3 be the length perpendiculars from the points (m2, 2m), (mm1, m+m1) and (m12, 2m1)

respectively on the line xcosα + ysinα + αα

cos

sin2

=0 then p1, p

2, p

3 are in

a. AP b. GP c. HP d. none of these9. If f(+y) =f(x)+f(y), ∀ x, y ε R and f(1)=2, then area enclosed by 3⏐x⏐+2⏐y⏐≤8 is

a. f(4) sq. u b. ⎟

⎠

⎞

⎜

⎝

⎛

2

1f(16) sq. u c.

3

1f(6) sq. u d.

3

1f(5) sq. u

10. The medians AD and BE of the triangle with vertices A(o, b), B(o, o) and C(a, o) are mutuallyperpendicular if

a. b= 2 a b. c. b=– 2 a d. a=5 2 b

11. The graph of the function y=cosxcos(x+2)–cos2(x+1) isa. a straight line passing through (0, –sin21) with slope 2.b. a straight line passing through (0, 0)c. a parabola with vertex (1, –sin21)

d. a straight line passing through the point ⎟

⎠

⎞

⎜

⎝

⎛ π1sin,–

22

parallel to the x-axis.

12. If line 2x+7y–1=0 intersect the lines L1=3x+4y+1=0 L

2=6x+8y–3=0 in A and B respectively,

then equation of a line parallel to L1 and L

2 and passes through a point P such that AP:PB=2:1

(internally) is (P is on the line 2x+7y–1=0).a. 9x+12y+3=0 b. 9x+12y–3=0 c. 9x+12y–2=0 d. none of these

13. The equation of a line through the point (1, 2) whose distance from the point (3, 1) has thegreatest possible value isa. y=x b. y=2x c. y=–2x d. y=–x

14. If the point (cosθ, sinθ) does not fall in that angle between the lines y=⏐x–1⏐ in which theorigin lies, the θ belongs to

a. ⎟

⎠

⎞

⎜

⎝

⎛ ππ2

3,

2b. ⎟

⎠

⎞

⎜

⎝

⎛ ππ2

,2

– c. (0, π) d. none of these

15. A line of fixed length (a+b) moves so that its ends are always on two fixed perpendicularlines. The locus of the point which divides this line into portion of lengths a and b isa. a circle b. an ellipse c. a hyper bola d. none of these

Numerical Grid Based problemsSolve the following problems and mark your response against their respective grids. Writeyour answer in the top row of the grid and darken the concerned numbers in the respectivecolumns.

ba 2=

43

16. The locus of the moving point P such that 2PA=3PB Where A is (0, 0) and B is (4, –3) is 5x2+5y2–72x+λy+μ=0 then the value of λ+μ must be.

17. If (λ,2) is an interior point of ΔABC formed by x+y=4, 3x–7y=8 and 4x–y=31 then λε(a, b), thevalue of 6a+8b must be.

18. A variable straight line through the point of intersection of the lines 4

y

3

x + =1 and 3

y

4

x + =1

meets the coordinate axes in A and B. If locus of the mid point of AB is 2xy=k(x+y) then thevalue of 343 k must be

Assertion and Reason (for θ 9-21)a. If Assertion is true and reasoing is also true, and reasoning is a correct explanation for assertion.b. If Assertion is True Reasning is also true but reasoing is not correct explanation for Assertion.c. Assertion is true reasoning is false Assertion is false is true.19. Assertion (A) : If centroid and circumcenter by a triangle are known, then its orthocenter and nine

point centre can be found.Reason (R) : Orthocenter, nine point centre, centroid and Circumcenter are collinear.a. A b. B c. C d. D

20. Assertion (A) : The line 2x+y+6=0 is perpendicular to the line x–2y+5=0 and second linepasses through (1, 3)Reason (R) : Product of the slopes of the lines is equal to –1.a. A b. B c. C d. D

21. Assertion (A) : The incenter of the triangle formed by the lines xcos ⎟

⎠

⎞

⎜

⎝

⎛ π9 +ysin ⎟

⎠

⎞

⎜

⎝

⎛ π9 –π=0, x

cos ⎟

⎠

⎞

⎜

⎝

⎛ π9

–ysin ⎟

⎠

⎞

⎜

⎝

⎛ π9

+π=0 and x cos ⎟

⎠

⎞

⎜

⎝

⎛ π9

4+ y sin ⎟

⎠

⎞

⎜

⎝

⎛ π9

4+π=0 is (0, 0)

Reason (R) : The point (0, 0) is equidistant from the three vertices of the triangle formed by

the lines xcos9

π+ysin

9

π–π=0, x cos

9

π–ysin

9

π+π=0, xcos ⎟

⎠

⎞

⎜

⎝

⎛ π9

4+ysin ⎟

⎠

⎞

⎜

⎝

⎛ π9

4+π=0

a. A b. B c. C d. D

ANSWERS

1. c 2. a 3. a 4. a 5. c 6. a7. a 8. b 9. c 10. b 11. d 12. c13. b 14. b 15. a 16. λ=54 μ=225, λ+μ =54+225=279

17.3

22< λ <

4

33 a =

3

22, b =

4

3318. Κ =

7

12 343Κ = 343 ×

7

12 =49 ×12=588

[6a + 8b = 6 ×3

22+8×

4

33=44+66]

= 44 + 66= 110

19. d 20. a 21. b

α

45

CIRCLE - IEquation of circles

BASIC CONCEPTS

Circle : A circle is the locus of points which are equidistant from a fixed point andlies on the same plane.

Fixed point is called centre of a circle and constant distance is called radius of thecircle

STANDARD EQUATION OF A CIRCLE

The equation of a circle with the centre at (h,k) and radiaus r is

(x–h)2 + (y–k)2 = r2

If centre is at the origin and radius is r then the equation of circle is x2+y2

=r2

GENERAL EQUATION OF A CIRCLE

x2+y2+2gx+2fy+c = 0 where g, f, and c are constants

centre (–g,–f) and radius is c–fg 22 +

CONDITIONS FOR A SECOND DEGREE EQUATION TO REPRESENT A CIRCLE

ax2+2hxy+by2+2gx+2fy+c = 0 is a second degree equation

(i) coefficient of x2 = coefficient of y2. ie., a = b

(ii) coefficient of xy = 0 ie., h = 0

If g2+f2–c>0 then the circle represents real circle with centre (–g, –f)

If g2 +f2 – c = 0 then the circle represents point circle since radius is zero

If g2+f2–c<0 then the circle is imaginary circle .

EQUATION OF CIRCLE IN VARIOUS FORMS

1 Equation of circle with centre (h.k) and passes through origin. is x2+y2+2hx+2ky = 0

Note that when a circle passes through origin the constant term must be zero

C(h,k)

xo

y

r

o

P

r

c(h,k)

P(x,y)

46

2 If the circle touches x–axis then its equation is (x ± h)2 +(y ± k)2 = k2 (or) x2+y2 ± 2hx ± 2ky+h2=0.In this case radius is ordinate of centre of a circle. Four circles possible

x

y

o

3 If the circle touches y–axis then its equation is (x ± h)2 + (y ± k)2 = h2 (or) x2+y2 ± 2hx ± 2ky+k2

=0. Here radius of the circle is abscissa of the centre. Four circles possible.y

xo

4 If the circle touches both the axes then its equation is (x ± r)2+(y ± r)2 =r2. Four circles possible

x2+y2 ± 2rx ± 2ry+r2 = 0

x

y

o

5 If the circle touches x–axis at origin then its equation is x2+(y ± k)2 = k2

x2+y2 ± 2ky = 0

x

y

47

6 If the circle touches y–axis at origin then its equation is (x ± h)2+y2 = h2 (or) x2+y2 ± 2hx = 0

y

xo

7 If the circle passes through origin and cuts intercepts a and b on the axes, then the equation ofcircle is x2+y2–ax–by = 0 and centre is c(a/2, b/2) four circles possible.

a

b

y

x

(a/2, b/2)

8 Equation of Circle on a given Diameter

If (x1,y

1) and (x

2,y

2) are end points of the diameter then the equation of circle is

(x–x1)(x–x

2) + (y–y

1)(y–y

2) = 0

CC

P(x,y)

A(x ,y )11

B(x ,y )22

9 Parametric form of Circle

x = h+rcos θy = k+rsinθWhere θ is parameter (0 ≤ θ ≤ 2 π)

In particular coordinates of any point on the circle x2+y2 = r2 is (rcos θ , rsin θ ) on the circle

x2+y2+2gx+2fy+c =0 is ( )( ))(sinc–fg,–ƒcosc–fgg– 2222 θ++θ++ ��

C(h,k)θr

P(x,y)

xo

y

48

10 Intercept made on the axes by a circle

Let the equation of circle is x2+y2+2gx+2fy+c = 0

AB = x - intercept = c–g2 2

CD = y - intercept = c–f2 2

y

x

(-g,-f)

MA B

C

D

N

0

49

CIRCLE - IIPosition of a Point

1. Position of a point with respect to a circle .

let the circle is x2+y2+2gx+2fy+c = 0

Point P(x1, y

1) lies outside,on or inside the circle

accordingly CP>, = , < radius

(or) S1 = x

12+y

12+2gx

1+2fy

1+c>, = , < 0

2 Maximum and Minimum Distance of a point from the circle

Let the circle x2+y2+2gx+2fy+c = 0 and point P(x1, y

1)

The maximum and minimum distance form P(x1, y

1) to the

circle are

PB = CB +CP

= r + CP

PA = |CP–CA| = |PC–r|

PB is maximum distance and PA is minimum distance.

EXAMPLES

1 If the equation px2+(3–q) xy +2y2–6q x+30y+6q = 0 represents a circle, then the values of p andq are

(a) 3,1 (b) 2,2 (c) 2,3 (d) 3,4

Solution :

ax2+by2+2hxy+2gx+2fy+c 0

represents a circle if h = 0 and a = b

∴ p = 2 and 3 – q = 0 ⇒ q = 3

correct option is (c)

2 The number if integral values of λ for which x2+y2+(1–λ )x+λy+5 =0 is the equation of a circlewhose radius cannot exceed 5 is

(a) 20 (b) 16 (c) 18 (d) 24

Solution :

radius of the equation c–fg 22 +

g = 2

–1 λ, f =

2

λc = 5

55–42

–1 22

≤λ+⎟

⎠

⎞

⎜

⎝

⎛ λ��

1+ λ 2–2 λ +λ 2–20 ≤ 100

2λ 2–2 λ –119 ≤ 0

C

P

P

P(x ,y )1 1

CP(x ,y )1 1AB

50

D = 4+8×119

= 4 +952

= 956

= 4(239)

λ = 4

23922 ±

= 2

239–1

2

2391 +

∴2

239–1 ≤ λ ≤

2

2391 +

–7.2 ≤ λ ≤ 8.2 (approx)

λ = –7, –6, –5,–4, ........ 8

number of values of λ is 16

3 The equation of the circle which passes through (1, 0) and (0, 1) and has its radius as small aspossible is

(a) x2+y2–2x–2y+1 = 0 (b) x2+y2–x–y = 0

(c) 2x2+2y2–3x–3y+1 = 0 (d) x2+y2–3x–3y+2 = 0

Solution :

The radius will be minimum, if the given points are the end points of a diameter. Then the equationof the circle is (x–1) (x–0) + (y–0)(y–1) = 0 ⇒x2+y2–x–y = 0

4 The centre of the circle x = –1+ 2cos θ , y = 3+2sin θ is

(a) (1,–3) (b) (–1, 3) (c) (1, 3) (d) None of these

Solution :

Rewrite the given equation

x+1 =2cos θ y–3 = 2sinθ

2

1x + =cosθ 2

3–y = sinθ

squaring and adding

2

2

1x⎟

⎠

⎞

⎜

⎝

⎛ + +

2

2

3–y⎟

⎠

⎞

⎜

⎝

⎛

= cos2 θ + sin2 θ

(x+1)2 + (y–3)2 = 4

centre (–1, 3) radius 2.

51

CIRCLE - IIIProblem Solving

1. The number of points (x,y) having integral coordinates satisfying the condition x2+y2<25 is

(a) 90 (b) 81 (c) 80 (d) 69

Solution :

Since x2+y2<25 and x and y are integers, the possible values of x & y ∈ (0, ± 1, ± 2, ± 3, ± 4),thus x and y can be chosen in 9 ways each and (x

1 y) can be 9×9 = 81 ways. But ( ± 3, ± 4)

( ± 4, ± 3) ( ± 4, ± 4) does not satisfy so we must exclude these points 3×4 = 12 ways.

Hence the number of permissible values are 81–12 = 69.

2. A point P moves in such a way that the ratio of its distance form two coplanar points is always afixed number ( ≠ 1) then its locus is(a) Straight line (b) circle (c) Parabola (d) a pair of Straight lines

Solution :Let two coplanar points be (0, 0) and (a, 0) according to the question we get

λ=BP

AP(constant)

( )λ=

+

+22

22

ya–x

yx

x2+y2 = λ 2(x2+y2–2ax+a2)

x2+y2+ 2

2

–1 λλ

(2ax–a2) = 0

represents equation of a circle3. The greatest distance of the point P (10, 7) from the circle

x2+y2–4x–2y–20 = 0 is(a) 10 (b) 15 (c) 5 (d) None of these

Solution :

Given equation of circle in x2+y2–4x–2y–20=0

S1 = 102+72–40–14–20>0

∴P(10, 7) lies outside the circle.

PB = PC+CB

= 22 68r ++= r +10

= 102014 +++= 5+10

= 15

C (2,1)P(10,7)

AB

52

4. If one end of a diameter of the circle x2+y2–4x–6y+11 = 0 be (3, 4), then the other end is

(a) (0, 0) (b) (1, 1) (c) (1,2) (d) (2,1).

Solution :

Centre of the circle is (2,3) one end of the diameteris (3,4). Since centre is the mid point of diameter

∴ 22

3 =+α & 3

2

4 =+β

α = 1, β = 2

∴ Other end points is (1,2).

5. ƒ(x,y) = x2+y2+2ax+2by+c = 0 represents a circle . If ƒ(x,0) = 0 has equal roots each being 2 andƒ(0,y) = 0 has 2 and 3 as its roots then the centre of circle is

(a) (2,5/2) (b) date are not sufficient

(c) (–2,–5/2) (d) data are inconsistent

Solution :

ƒ(x,0) = 0 ⇒ x2+2ax+c = 0 = (x–2)2

∴ c = 4. & 2a = –4. ⇒ a = – 2

ƒ(0,y) = 0 ⇒ y2+2by+c = 0 = (y–2) (y–3)

= y2–5y+6

∴ b = –5/2 & c = 6

∴c is not unique so data are inconsistent

6. If p and q are the largest distance and the shortest distance respectively of the point (–7,2) from

any point ( α ,β ) on the curve whose equation is x2+y2–10x–14y–51 = 0 then G.M of p and q isequal to

(a) 2 11 (b) 5 5 (c) 13 (d) None of these

Solution :

The centre C of the circle is (5,7) and the radius is 514925 ++ = 55125 =

PC = 22 512 + = 169 = 13

∴p = PB = PC+CB q = PA = PC–CA

= 13+ 55 = 13– 55

GM of p&q = pq = ( )( )55–135513+

= ( )22 55–13

= 125–169

C (2,3)AB

, β)

T(2,0)

P(2,4)

A

B

C (5,7)P

AB

53

= 44

= 112

EXAMPLES

7. If (1+ α x)n = 1+8x+24x2+................. and a line through P( α ,n)cuts the circle x2+y2 = 4 in A and B, then PA.PB is

(a) 4 (b) 8 (c) 16 (d) 32.

Solution :(1+ α x)n = 1+8x+24x2+.................

1+n α x+ 2C )x(n

2α +.................= 1+8x+24x2+....................

n α x = 8x 2C )x(n

2α = 24x2

n α = 82

)1–n(n α 2= 24

2

)–n(n ααα = 24

2

)–8( α = 3

⇒ 8– α = 6

α = 2

∴ n = 4.

P( α ,n) = P(2,4) and PT is a tangent of length 4

We know that

PT2 = PA.PB = 42 = 16

(Secant tangent theorem)

8. If ƒ(x+y) = ƒ(x)ƒ(y) ∀ x, y, ƒ(1) = 2 and α n = f(n), n∈N then the equation of the circle having

( α 1, α 2

) and ( α 3, α 4

) as the ends of its one diameter is

(a) (x–2)(x–8)+(y–4)(y–16) = 0 (b) (x–4)(x–8)+(y–2)(y–16) = 0

(c) (x–2)(x–16)+(y–4)(y–8) = 0 (d) (x–6)(x–8)+(y–5)(y–6) = 0

Solution :

ƒ(x+y)=ƒ(x)ƒ(y)

ƒ(2)= ƒ(1+1) = ƒ(1)ƒ(1) = 22 = α 2

α 3=ƒ(3)=23, ƒ(4) = 24= α 4

( α 1, α 2

)= (2,4) & ( α 3, α 4

) = (8, 16)

Equation of circle in diameter form

(x–x1) (x–x

2) + (y–y

1)(y–y

2) = 0

(x–2) (x–8) + (y–4)(y–16) = 0

T(2,0)

P(2,4)

A

B

54

9. If A and B are two points on the circle x2+y2–4x+6y–3 = 0 which are farthest and nearest respec-tively from the point (7, 2) then

(a) A ≡ (2–2 2 , –3–2 2 ) (b) A ≡ (2+2 2 , –3+2 2 )

(c) β ≡ (2+2 2 , –3+2 2 ) (d) β ≡ (2–2 2 , –3+2 2 )

Solution :

x2+y2–4x+6y–3 = 0

Centre of the circle is (2, –3)

Radius of the circle is = 332 32 ++ = 16 = 4

S1 = 49+ 4–28+12–3

= 65–31

= 34>0

∴point (7, 2) lies outside the circle

PC = ( ) ( )22 322–7 ++

= 22 55 +

= 2550 =CA = CB = r = 4

Farthest point PA = PC+CA

= 5 42 +Nearest point PB = PC–CB

= 4–25

By tinding Point of pnteraction

∴The coordinates of A and B are

A ( ) ( )( )22–3–,22–2 and B ( )223,–222 ++

PRACTICE QUESTIONS

1. The points A and B in a plane are such that for all points P lies on circle Satisfying PB

PA = k, then k

will not be equal to(a) 0 (b) 1 (c) 2 (d) None of these

2. If the line hx+ky = 1 touches x2+y2 = a2 then the locus of the point (h,k) is a circle of radius

(a) a (b) 1/a (c) a (d)a

1

3. Equation of incircle of equilateral triangle ABC where B(2,0), C(4,0) and A lies in the fourth

C (2,-3)P(7,2)

BA

55

quadrant is

(a) x2+y2–6x+3

y2+9 = 0 (b) x2+y2–6x–

3

y2+9 = 0

(c) x2+y2+6x+3

y2+9 = 0 (d) None of these

4. A variable circle having chord of radius ‘a’ passes through origin meets the coordinate axes inpoints A and B. locus of centroid of triangle OAB, ‘O’ being the origin is

(a) 9(x2+y2) = 4a2 (b) 9(x2+y2) = a2

(c) 9(x2+y2) = 2a2 (d) 9(x2+y2) = 8a2

5 The locus of the centre of the circle which cuts a chord of length 2a from the positive x–axis andpasses through a point on positive y–axis distant b from the origin is

(a) x2+2by = b2+a2 (b) x2–2by = b2+a2

(c) x2+2by = a2–b2 (d) x2–2by = b2–a2

6 The number of circle having radius 5 and passing through the points (–2,0) and (4,0) is

(a) one (b) two (c) four (d) infinite

7 The equation of the smallest circle passing through the intersection of the line x+y=1 and the circlex2+y2 = 9 is

(a) x2+y2+x+y–8 = 0 (b) x2+y2–x–y–8 = 0

(c) x2+y2–x–y+8 = 0 (d) None of these

8 The number of the points on the circle x2+y2–4x–10y+13=0 which are at a distance 1 from thepoint (–3,2) is

(a) 1 (b) 2 (c) 3 (d) None of these

9 The locus of the mid-point of a chord of the circle x2+y2 = 4 which subtends a right angle at theorigin is

(a) x+y=2 (b) x2+y2=1 (c) x2+y2=2 (d) x+y = 1

10 The area of the triangle formed by joining the origin to the points of intersection of the line x 5 +2y

=3 5 and circle x2+y2 = 10 is

(a) 3 (b) 4 (c) 5 (d) 6

11 If (–3,2) lies on the circle x2+y2+2gx+2fy+c=0 which is concentric with the circle x2+y2+6x+8y–5 = 0 then c is

(a) 11 (b) –11 (c) 24 (d) None of these

12 A variable circle passes through the fixed point A(p,q) and touches x–axis. The locus of the otherend of the diameter through A is

(a) (x–p)2=4qy (b) (x–q)2=4qy (c) (y–p)2=4qx (d) (y–q)2=4px

13 If the lines 2x+3y+1 = 0 and 3x–y–4 = 0 lie along diameters of a circle of circmference 10 π , then

56

the equation of the circle is

(a) x2+y2–2x+2y–23=0 (b) x2+y2+2x+2y–23=0

(c) x2+y2–2x–2y–23=0 (d) x2+y2+2x–2y–23=0

14 The lines 2x–3y=5 and 3x–4y = 7 are diameters of a circle having area as 154 sq.unit, then theequation of the circle is

(a) x2+y2+2x–2y=62 (b) x2+y2+2x–2y=47

(c) x2+y2–2x+2y=47 (d) x2+y2–2x+2y=62

15 The equation of circle which passes through the origin and cuts off intercepts 5 and 6 from thepositive parts of the axes respectively, is

(a) (x+5/2)2+(y+3)2=4

61(b) (x–5/2)2+(y–3)2=

4

61

(c) (x–5/2)2–(y–3)2=4

61(d) (x–5/2)2+(y+3)2=

4

61

ANSWERS

1. b 2. b 3.a 4. a 5.c 6. b

7.b 8. d 9. c 10. c 11. b 12. a

13. a 14. c 15. b

57

CIRCLE - IVTangents and Normals

Topics covered :1. Distance of a line from a circle2. Different forms of the equations of tangents3. Length of tangent from a point to a circle.4. Normal to a circle at a given point.

1. Line and a circle

Let S = 0 and L = 0 be a circle and a line. If r is the radius of the circle and d is the length ofperpendicular from the centre on the line then,

(i) d > r ⇒ line does not meet circle

L=0

d

CS=0

(ii) d = r ⇒ line touches the circle. It is a tangent to the circle

L=0

d

CS=0

r

(iii) d < r ⇒ line intersect the circle line is a secant to the circle

L=0

d

CS=0

r

(iv) d = 0 ⇒ line is a diameter of the circle L=0

C

S=0

If y = mx+c is a line and x2+y2 =r2 is a circle then

(i) c2 > r2 (1+m2) ⇒The line is a secant of the circle.The line intersects the circle in twodistinct points.

(ii) c2 = r2 (1+m2) ⇒The line is a tangent to the circle.The line touches the circle at uniquepoint.

(iii) c2 < r2 (1+m2) ⇒The line does not meet the circle

58

Equations of Tangents1 Point form : Equation of the tangent to the circle x2+y2 = a2 at the point (x

1, y

1) on it is :

xx1+yy

1=a2

2 Equation of the tangent to the circle x2+y2+2gx+2fy+c=0 at the point (x1, y

1) on it is

xx1+yy

1+g(x+x

1)+ƒ(y+y

1)+c = 0

I. Parametric formsEquation of the tangent to the circle x2+y2 = a2 at the point (a cosθ ,asinθ ) on it its xcosθ +ysinθ= a

II. Slope form

The equation of a tangent of slope m to the circle x2+y2 = a2 is y = mx ± a 2m1+

The coordinates of the point of contact are ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

++

+±

22 m1

a,

m1

am

(i) Condition for a line y = mx+c to be a tangent to the circle x2+y2=a2 is c2 =a2(1+m2) or c =

± a 2m1+(ii) Condition that the line x+my+n = 0 touches the circle x2+y2+2gx+2fy+c = 0 is ( g+mf+n)2

= ( 2+m2)(g2+ƒ2–c)(iii) Equation of tangent to the circle x2+y2+2gx+2fy+c = 0 in terms of slope is y = mx+mg–

f ± c–ƒg 22 + 2m1+(iv) The line x+my+n = 0 touches the circle (x–a)2+(y–b)2 = r2 if (a +bm+n)2 = r2( 2+m2)(v) If the line y = mx+c is the tangent to the circle x2+y2 = r2 then point of contact is given by

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

c

r,

c

mr– 22

(vi) If the line ax+by+c = 0 is the tangent to the circle x2+y2 = r2 then point of contact is given

by ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

cbr–

,car– 22

III. Tangents from a point outside the circleIf circle is x2+y2 = a2 andany tangent to the circle is

y = mx+a 2m1+P(x

1, y

1) lies on the tangent

∴ y1 = mx

1+a 2m1+

y1–mx

1 = a 2m1+

Squaring(y

1–mx

1)2 = a2(1+m2)

m2(x12–a2) – 2mx

1y

1+y

12–a2 = 0 is a quadratic equation

in m which gives 2 values of m.We get two equations of tangents.

C (0,0) P(x ,y )1 1

a

A

B

59

Tangents from a point outside the circleLet the circle be x2+y2=r2 and a point P(x

1,y

1) outside

the circle .Let the slope of the tangent is m then equation of thetangent is y–y

1 = m (x–x

1)___________(1)

Now find the distance of this line from the centre (0,0)and equate to the radius. We get equation in m solve andget two values of m and substitute in (1)

IV Length of the tangent from a point to a circle.Let the circle be S≡ x2+y2+2gx+2ƒy+c = 0 thencentre and radius of circle are (–g, –ƒ) and

cg –ƒ22 + respectively and let P(x1, y

1) be any

point outside the circle.

PT = cyfgxyx +++++ 1112

12

1 22

= 1S

V. Power of point with respect to a circleThe power of P(x

1, y

1) with respect to S ≡

x2+y2+2gx+2ƒy+c = 0 is equal to PA. PB orPC.PD which is x

12+y

12+2gx

1+2ƒy

1+c = 0 ⇒

S1=0 (: PA.PB = PC.PD= PT2)

Note :(i) The power of the point outside the circle is posi-tive(ii) The power of the point on the circle is zero(iii) The power of the point inside the circle is negative

VI. Pair of tangentsThe equation of the pair of tangents drawn from the pointP(x

1,y

1) to the circle S = 0 is SS

1 = T2

Where S = x2+y2+2gx+2ƒy+cS

1 = x

12+y

12+2gx

1+2ƒy

1+c

T = xx1+yy

1+g(x+x

1)+ƒ(y+y

1)+c

Note : The pair of tangents from (0,0) to the circlex2+y2+2gx+2ƒy+c = 0 are at right angles if g2+f2 = 2c

VII. Normal to a circle at a given pointThe normal of a circle at any point is a straight line which is perpendicular to the tangent at thepoint and always passes through the centre of the circle.(1) Point formTo find the equation of normal to the circle x2+y2 = a2 at the point p(x

1,y

1) on it .

Since we know that normal passes through the centre of a circle . So we get two points on normalusing two point form of a line we get the equation of normal as

P(x ,y )1 1

T

A

B

C

D

P(x ,y )1 1

S=0

B

A

C (-g,-f) P(x,y)1 1

T

c fg 22 + −

60

0–x

0–x

0–y

0–y

11

=

or

11 x

x

y

y =

xy1–yx

1 = 0

To find normal at (x1, y

1) of second degree conics ax2+2hxy+by2+2gx+2ƒy+c = 0__________(1)

then according to determinant efg

fbh

gha

Write first two rows as ax1+hy

1+g and hx

1+by

1+f

Then normal at (x1,y

1) of (1) is ghyax

x–x

11

1

++ = fbyhx

y–y

11

1

++

� If equation of circle is x2+y2 = a2

here a= b = 1 and h = 0 = g = f

∴1

1

x

x–x =

1

1

y

y–y

1x

x –1 =

1y

y –1

1x

x =

1y

y is equation of normal at (x

1, y

1)

� If equation of circle is x2+y2+2gx+2fy+c = 0

here a = b = 1 and h = 0

Then gx

x–x

1

1

+ = fy

y–y

1

1

+

C (0,0)

a

C (0,0)

P(x ,y )1 1

a

T

61

CIRCLE - VTangents and Normals

1. Chord of contactLet the equation of circle be x2+y2 = r2 PA andPB are pair of tangents drawn from the pointP(x

1,y

1) then AB is the chord of contact with A

and B as its points of contact.∴Equation of chord of contact AB is xx

1+yy

1=a2

Equation of chord of contact look like equationof tangent at point but point are differentIf the equation of circle be x2+y2+2gx+2fy+c =0 then the equation of chord of contact isxx

1+yy

1+g(x+x

1)+f(y+y

1)+c=0

2. Equation of the chord bisected at a given pointLet the equation of circle be x2+y2=r2 and AB is a chord of it Let (M(x

1,y

1) be midpoint of AB.

Slope of CM = 1

1

x

y

Slope of AB = 1

1

y

x–

Equation of chord AB is

y–y1=

1

1

y

x–(x–x

1)

yy1–y

12 = – xx

1+x

12

xx1+yy

1=x

12+y

12

xx1+yy

1–a2=x

12+y

12–a2

T = S1

If the equation of circle be x2+y2+2gx+2fy+c = 0 then the equation of chord which is bisected at(x

1, y

1) is

`xx1+yy

1+g(x+x

1)+f(y+y

1)+c = x

12+y

12+2gx

1+2fy

1+c

EXAMPLES :1 Find the equation of tangent to the circle x2+y2–2ax = 0 at the point (a(1+cos α ), a sin α )Solution :

Equation of tangent of x2+y2–2ax = 0at (a(1+cos α ),asin α ) isax(1+cos α )+aysin α –a(x+a(1+cos α )) = 0ax+axcos α +aysin α –ax–a2(1+cos α ) = 0axcos α +aysin α =a2(1+cosα )xcos α +ysin α = a(1+cos α )

2 Find the equations of the tangents to the circle x2+y2 = 9 which make an angle of 60° with theaxis.

P(x ,y )1 1

A

B

P(x ,y )1 1

A

B

C (0,0)C (0,0)

M(x y )1, 1

A

B

62

Solution :Since tangents make an angle of 60° with the x–axis so slope of tangent

m = tan60°=m= 3radius of circle x2+y2 = 9 is 3we know equation of tangent to a circle x2+y2 = a2 is

y = mx ± a 2m1+

∴ y = 3 x ± 3 31+

= 3 x ± 6

or 3 x–y ± 6 = 0 ie., 3 x–y+6 = 0 and 3 x–y–6 =0 are equations of tangents.

3 Show that the line (x–2) cosθ +(y–2) sinθ = 1 touches a circle for all values of θ . Find the circle.Solution :

Since the line (x–2) cosθ +(y–2) sinθ = 1_______(1) touches a circle so it is a tangent equationto a circle.Equation of tangent to a circle at (x

1,y

1) is (x–h)x

1+(y–k)y

1 = a2 to a circle (x–h)2+(y–k)2= a2

comparing (1) and (2) we getx–h = x–2 y–k = y–2 and a2 = 1x

1 = 1cos θ y

1 = 1sin θ

∴Required equation of circle is(x–2)2 + (y–2)2 = 1x2+y2–4x–4y+7 = 0

4 Find the equation of the normal to the circle x2+y2 =2x which is parallel to the line x+2y = 3Solution :

Equation of normal at (x1,y

1) of x2+y2 –2x = 0 is

1–x

x–x

1

1 = 0–y

y–y

1

1 ⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛

+=

+ fby

y–y

gax

x–x

1

1

1

1

Slope of this equation is 1–x

y

1

1

Slope of x+2y = 3 is 21–

Since given that normal is parallel to x+2y =3

∴ 1–x

y

1

1 =

2

1–

2y1=–x

1+1 therefore locus of (x

1, y

1) is x

1+2y

1 = 1

It is the equation of normal (x+2y)=15 Show that the line 3x–4y = 1 touches the circle x2+y2–2x+4y+1 = 0 find the coordinates of the

point of contact.Solution :

Centre and radius of circle x2+y2–2x+4y+1 = 0

⇒

63

is (1,–2) and 1–41+ = 2 respectively.

If the distance of a line 3x–4y = 1 from the centre (1,–2) is equal to radius then the line touches orit is tangent to a circle.

= 22 43

1–)2(–4–13

+×

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

=+

++d

ba

cbyax22

11

= 51–83 +

= 2∴ line 3x–4y= 1 touches the circle.Let point of contact be (x

1,y

1) then equation of tangent to a circle x2+y2–2x+4y+1=0 is

xx1+yy

1–(x+x

1)+2(y+y

1)+1 = 0_________________(1)

x(x1–1)+y(y

1+2)–x

1+2y

1+1 = 0__________________(2)

and given line 3x–4y–1 = 0(1) and (2) are idenfical then comparing (1) and (2) we get

31–x1 =

4–2y1 +

= 1–

1y2x– 11 ++

–x1+1= –3x

1+6y

1+3 or 2x

1–6y

1–2=0

–y1–2=4x

1–8y

1–4 or 4x

1–7y

1–2=0

Solving these two equations of x1,y

1 we get x

1 =

5

1– and y

1 =

5

2–

∴ point of contact is ⎟

⎠

⎞

⎜

⎝

⎛

5

2–,

5

1–

6 The angle between a pair of tangents from a point P to the circlex2+y2+4x–6y+9sin2 α +13cos2 α = 0 is 2 α . Find the equation of the locus of the point P.

Solution :Let the coordinate of Pbe (x

1,y

1) and given circle is x2+y2+4x–6y+9sin2 α +13cos2 α =0

(x+2)2+(y–3)2–4–9+9sin2 α +13cos2 α = 0(x+2)2+(y–3)2–9sin2 α –13(1–cos2 α ) = 0(x+2)2+(y–3)2–9sin2 α –13sin2 α = 0(x+2)2+(y–3)2 = 4sin2 α = (2sin α )2

∴ centre and radius of circle is (–2,3) and 2 sin α respectivelyDistance of P and C is

PC = 21 )3–y()2x( 1

2 ++

In Δ PCR

sin α = 21

2 )3–y()2x(

sin2

1 ++α

C (-2,3)

P(x ,y )1 1

2sinα

R

P(x ,y )1 1

T

αα2α

64

TP

R x +y=c22 2

x +y =b22 2

x +y =a22 2

o

or 21

2 )3–y()2x( 1 ++ = 2

Squaring(x

1+2)2+(y

1–3)2 = 4 or (x+2)2+(y–3)2 = 4

∴ locus of point P is a circle7 If the length of tangent from (f,g) to the circle x2+y2 =6 be twice the length of the tangent from (f,g)

to circle x2+y2+3x+3y = 0 then prove that f2+g2+4f+4g+2 = 0Solution :

According to the question

6–gf 22 + = 2 g3f3gf 22 +++Squaring both sidef2+g2–6 = 4 (f2+g2+3f+3g)3f2+3g2+12f+12g+6 = 0Divide by 3 we get f2+g2+4f+4g+2 = 0

8 The chord of contact of tangents drawn from a point on the circle x2+y2 = a2 to the circle x2+y2 =b2 touches the circle x2+y2 = c2 show that a,b,c are in GP.

Solution :Let P (a cosθ ,asinθ ) be a point on the circle x2+y2 = a2 Then equation of chord of contact to the

circle x2+y2 = b2 from P (a cos θ , asin θ ) is

x(a cos θ )+y (asin θ ) = b2

axsinθ +aysinθ = b2

It is a tangent to the circle x2+y2= c2

∴ length of perpendicular to the line = radius.

2

2

a

b– = c

b2 = ac∴ a, b.c are in G.P.

65

CIRCLE - VITangents and Normals

1 Director circle and its equationThe locus of the point of intersection of two perpendicular tangents to a given circle is known asits director circle.Equation of Director Circle

C(0, 0)

Br r

A

P(h,k)

90°

45° 45°

Let P(h,k) be the point of intersection of tangents to a circle x2+y2 = r2at right angle.ACBP is a square

∴AC = CPsin45°

r = 2

kh 22 +

Squaring we get2r2 = h2+k2

or x2+y2 = 2r2 is the required equation. of director circle2 Intersection of two circles, common Tangents to two circles

Let the two circles be (x–g1)2+(y–f

1)2 = r

12 and (x–g

2)2+(y–f

2)2 = r

22

with centres C1(g

1,f

1) and C

2(g

2,f

2) and radii r

1 and r

2 respectively

Different cases of intersection of two circlesCase I When |C

1C

2|>r

1+r

2

ie., distance between the centre is greater than sum of the radii.In this case there are 4 common tangents can be drawn.Two direct common tangents (circles lies on the same side of the tangent)ABE and CDE

I

G

C1

r1

r2

C2

C

A

B

FJ D

EH

Two indirect (Transverse) common tangents (circles lies opposite side of the tangent)GFH & IFJ.Note that centres of two circles and point of intersection of tangents are collinear also

66

EC

EC

2

1 =

2

1

r

r & FC

FC

2

1 =

2

1

r

r

To find the equations of common tangents.Let us assume equation of tangent of any circle in slope form be

(y+f) = m(x+g)+r 2m1+Points E & F satisfy this equation. Substitute the coordinates of E & F to get the values of m.Substitute the values of m we get required respective common tangent equationCase II When|C

1C

2| = r

1+r

2

D

G

H

C1C2

C

A

B

E

ie., distance between the centre is equal to sum of the radii.In this case there are 3 common tangents 2 direct common tangent and one transverse commontangentThe equation of tangent at point F is S

1–S

2 = 0 where S

1 = 0 and S

2 = 0 are equations of circles.

Coordinate of F are ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

++

++

21

1221

21

1221

rr

frfr,

rr

grgr

Case III When |C1C

2|<r

1+r

2

D

C1C2

C

A

B

E

ie,. Distance between the centre is less than sum of the radii.In this case only two direct common tangents.Case IV When |C

1C

2| = |r

1–r

2|

C1

S2

S1

C2

F

67

ie., distance between the centre is equal to difference of the radii.Then the two circles touch each other internally.In this case only one direct common tangent.Equation of common tangent is S

1–S

2 = 0

F divides line joining C1 and C

2 externally in the ratio r

1:r

2

∴coordinates of F are ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

21

1221

21

1221

r–r

fr–fr,

r–r

gr–gr

Case V When |C1C

2|<|r

1–r

2|

C1

S2

S1

C2

ie., distance between the centre is less than the difference of the radii. Then one circle contains theotherIn this case there is no real Common tangents.

68

3 Length of direct common tangent if |C1C2|>r1+r2

length of direct common tangent = 221

2 )r–r(–d

length of transverse common tangent = 221

2 )rr(–d +Where d = |C

1C

2| and r

1, r

2 are radii of the circles

4 Pole and Polar of the circleLet P(x

1,y

1) be any point inside or outside the circle. Draw chords AB and CD passing through P

. If tangents to the circle at A and B meet at Q (h,k) then locus of Q is called the polar of P withrespect to circle and P is called the pole and if tangents to the circle atC and D meet at R, then the straight line QR is polar with P as its pole.Let the equation of circle be x2+y2 = a2 then AB is a chord of contact of Q(h,k)∴xh+yk = a2 is its equationP (x

1,y

1) lies on AB

∴hx1+ky

1 = a2.

P(x ,y )pole1 1

Polar

DR

Q(h,k)

A

C

B

Hence locus of Q(h,k) is xx1+yy

1=a2 which is polar of P(x

1,y

1) with respect to circle x2+y2= a2

Equation of polar of the circle x2+y2+2gx+2fy+c = 0 with respect to (x1,y

1) is

xx1+yy

1+g(x+x

1)+f(y+y

1)+c = 0

Coordinates of pole of a lineLet the polar line ax+by +c = 0 with respect to the circle x2+y2 = r2

Let the pole be (x1,y

1) then equation of polar with respect to the circle x2+y2 = r2 is

xx1+yy

1 – r2 = 0____________________(1)

Which is same as ax+by+c =0______________(2)

comparing (1) and (2) we get

a

x1 = b

y1 =c

r– 2

x1 =

c

ar– 2

& y1 =

cbr– 2

69

∴The coordinate of pole is ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

cbr–

,car– 22

Properties of pole and polar(i) The distance of any two points P(x

1,y

1) and Q(x

2, y

2) from the centre of a circle is proportional to

the distance of each from the polar to the other(ii) If O be the centre of a circle and P any point then OP is perpendicular to the polar of P(iii) If O be the centre of a circle and P any point then if OP (produced if necessary) meet the polar

of P in S then OP.OS = r2

(iv) If the polar of P(x1,y

1) with respect to a circle passes through S(x

2,y

2) then the ploar of S will pass

through P and such points are said to be conjugate points(v) If the pole of the line ax+by+c=0 with respect to a circle lies on another line a

1x+b

1y+c

1 = 0.

Then the pole of the second line will lie on the first line and such lines are said to be conjugatelines.

Note :(i) P(x

1,y

1) and Q(x

2,y

2) are convent points w.r.t the circle x2+y2+2gx+2fy+c = 0 if

x1x

2+y

1y

2+g(x

1+x

2)+f(y

1+y

2)+c = 0

(ii) If P and Q are confined points w.r.t to a circle with centre at O and radius r, thenPQ2 = OP2+OQ2–2r2.

5 Common Chord of two circlesThe common chord joining the point of intersection of two circles is called their common chord .IfS = 0 and S1 = 0 be two interesting circles then the equation of their common chord is S–S1 = 0Let the equations of circle oneS = x2+y2+2g

1x+2f

1y+c

1 = 0

S1 = x2+y2+2g2x+2f

2y+c

2 = 0

Then equation of common chord AB isS–S1 = 0

C1C2

S=0

A

0S =′

2x(g1–g

2)+2y(f

1–f

2)+c

1 –c

2 = 0

Length of common chord is 2AM = 22 )MC(–)AC( 11

C1M is the length of perpendicular from the centre C

1 to common chord and C

1A is radius of

circle.Note :(i) Common chord AB becomes maximum length when it is a diameter of the smaller one.(ii) Circle on the common chord a diameter then centre of the circle passing through A and B lie on

the common chord of the two circle(iii) If the length of common chord is zero then the two circles touch each other and the common

m

70

chord be comes the common tangent to the two circles at the point of contact.Examples1 There are two circles whose equations are x2+y2 = 9 and x2+y2–8x–6y+n2 = O n∈Z. If two

circles have exactly two common tangents then the number of possible values of nSolution : For x2+y2 = 9 centre is (0,0) and radius 3 and for x2+y2–8x–6y+n2 = 0 centre is (4,3)

and radius 222 n34 ++ = 2n–25We know that to get exactly two common tangents the circles must intersect is|C

1C

2|<r

1+r

2

22 34 + < 3+ 2n–25

5< 3+ 2n–25

2< 2n–254< 25–n2

n2<21

or – 21 < n< 21 and 25–n2 ≥0

25 ≥ n2

or –5 ≤ n≤ 5n∈Z. so n = –4, –3, –2, –1, 0, 1, 2, 3, 4∴ number of possible value of n is 9

2 Find all the common tangents to the circles x2+y2–2x–6y+9 = 0 and x2+y2+6x–2y+1 = 0Solution : Centre and radius of circle x2+y2–2x–6y+9 = 0 is C

1(1,3) and r

1 = 1 respectively

Centre and radius of circle x2+y2+6x–2y+1 = 0 is C2(–3,1) and r

2 = 3

C2C

2= 22 24 + r

1+r

2 = 1+3 = 4

= 416 + ∴ |C1C

2|>r

1+r

2

= 20 ∴These are four common tangents

= 2 5 2 direct and 2 transverse.

To find equations of direct common tangentsSince the coordinate of E divides the line C

1C

2 in the ratio r

1:r

2 ie., 1:3 externally

∴coordinate of E is

⎟

⎠

⎞

⎜

⎝

⎛

3–13.3–1.1

,3–1

1.3–)3(–1 or E (3,4)

C

C2(–3,1)

A

1

3

C1(1,3)

B

D

E

71

Equation of AB through E (3,4) with slope m isy–4= m

1(x–3) or m

1x–y+(4–3m

1) = 0__________________(1)

If is a tangent to the circle so distance of the line from the centre is equal to radius

∴ 3 = 1m

m3–41–m3–2

1

11

+

+

3 = 1m

m6–32

1

1

+`

3 = 1m

m6–32

1

1

+

1m 21 + = 1–2m

1

Squaringm

12+1 = 1+4m

12–4m

1

3m12–4m

1 = 0 ⇒ m

1 = 0 or m

1 = 4/3

Substitute the values of m1 in equation (1) we get the equations of tangents

y = 4 and 4x–3y = 0To find the indirect or transverse common tangentsThe coordinates of T dividingC

1C

2 internally in the ratio

r1: r

2 = 1: 3 is

⎟

⎠

⎞

⎜

⎝

⎛ ++4

91,

433–

= ⎟

⎠

⎞

⎜

⎝

⎛

25

,0

Equation of line through T with slope m2 is

y –5/2 = m2(x–0) or 2m

2x–2y+5 = 0

P

C2(–3,1)C1(1,3)

R

TS

Q

Since it is a tangent to the circles

72

∴distance of the line from its centre is equal to radius.

∴ 4m4

5)1(2–)3(–m22

2

2

+

+ = 3

1m2

3m6–2

2

2

+

+ = 3

or 3–6m2 = 6 1m 2

2 +

1–2m2 = 2 1m 2

2 +Squaring1+4m

22–4m

2= 4m

22+4

4m2 = –3 or m

2 =

4

3 and ∞ (as coefficient of m

22 = 0)

∴The equations of transverse common tangents are3x+4y–10 = 0 and x = 0

3 If the circle C1 = x2+y2 = 16 intersects another circles C

2 of radius 5 in such a manner that the

common chord is of maximum length and has a slope equal to 3/4 the find the coordinates of thecentre C

2

Solution: ⎟

⎠

⎞

⎜

⎝

⎛

5

12–,

5

9 & ⎟

⎠

⎞

⎜

⎝

⎛

5

12,

5

9–

When two circles intersect, the common chord of maximum length will be the diameter of smallercircle.∴AB is diameter of smaller circleC

2A = 5 and C

1A = 4

C2(h,k)

5 43C1(0,0)

A

B

∴C1C

2 = 22 4–5

C1C

2 = 3

Given that slope of AB is 3/4∴slope of C

1C

2 is –4/3

73

and slope of C1C

2 is

h

k

∴ 3

4–=

h

k

k = 3

4–h

C1C

2 = 22 kh + = 3

or h2+k2 = 9__________________(1)

Substitute k = 3

4–h in equation (1)

h2+9

16h2 = 9

25h2 = 81

h2 = 25

81

h = ± 5

9∴K = ±

5

12

∴coordinates of C2 are ⎟

⎠

⎞

⎜

⎝

⎛

5

12–,

5

9 and ⎟

⎠

⎞

⎜

⎝

⎛

5

12,

5

9–

4 The circle x2+y2–4x–8y+16 = 0 rolls up the tangent to it at (2+ 3 ,3) by 2units, assuming the x-

axis as horizontal, find the equation of the circle in the new position.Solution ;Given circle is x2+y2–4x–8y+16 = 0

Equation of tangent to the circle at (2+ 3 ,3) is (2+ 3 )x+3y–2(2+ 3 +x)–4(y+3)+16 = 0

3 x–y–2 3 = 0

slope of this line tanθ = 3 ∴ θ = 600.

A and B are the centres ofThe circles in old and new positions respectively then

A(2,4) B

QP( )3,32 +

74

A (2,4) and B(2+2cos600), 4+2sin60) (⇒AB = 2)

∴B(3,4+ 3 )

Hence equation of circle at new position is

(x–3)2+(y–4– 3 )2 = 22

x2+y2–6x–2(4+ 3 )y+24+8 3 =0

5 The pole of a straight line with respect to the circle x2+y2=a2 lies on the circle x2+y2=9a2

Prove that the straight line touches the circle x2+y2 = 9a 2

Solution Let pole be (x1,y

1)

The equation of the polar with respect to x2+y2 = a2isxx

1+yy

1 = a2

But given that P(x1,y

1) lies on the circle x2+y2 = 9a2

∴x12+y

12 = 9a2

If straight line xx1+yy

1 = a2 touches the circle then distance of this line from the centre of the circle

x2+y2 = a2/9 must be a/3(radius)

ie., 21

21

2

yx

a–

+

21

21

2

yx

a

+ = 2

2

a9

a =

a3

a 2

= 3

a = radius

Hence straight line xx1+yy

1 = a2 touches the circle x2+y2 = a2/9

6 Prove that the polar of a given point with respect to any one of the circles x2+y2–2kx+c2 = 0,where k is a variable, always passes through a fixed point, whatever be the value of k.Solution: Let P(x

1,y

1) be pole, then equation of polar with respect to the circle x2+y2–2kx+c2 = 0

isxx

1+yy

1–k(x+x

1)+c2=0

(xx1+yy

1+c2)–k(x+x

1) = 0___________________(1)

L1+λL

2 = 0

Where L1 = xx

1+yy

1+c2 = 0 & L

2 = x+x

1 = 0

Hence equation (1) represents the family of lines passing through the point of intercection of twolines L

1 & L

2

Solving equation L1 = 0 & L

2 = 0 we get

x = –x1 & –x

12+yy

1+c2 = 0

y = 1

221

y

c–x

75

∴ The fixed point is

( )⎟⎟

⎟

⎠

⎞

⎜⎜

⎜

⎝

⎛

1

221

1 y

c–x,x– is independent of k.

PRACTICE QUESTIONS1 Passage 1

Let a straight line be drawn from a point P to meet the circle in Q and R. Let the tangents at Q andR meet at T. The locus of T is called the polar of P with respect to the circle . The given point P iscalled the pole of the polar lineLet P(x

1,y

1) be the given point lying outside the circle in fig (i) and inside the circle in fig (ii)

Through P, draw a line to meet the circle in Q and R. Let the tangents to the circle at Q and Rmeet in T (h,k).It is required to find the polar of P ie., locus of T.Equation of QR the chord of contact of the tangents draw from T to the circle x2+y2 = a2 is xh+yk= a2_______________(1)since (1) passes through P(x

1,y

1)

∴x1h+y

1k = a2

p(x ,y )1 1

o

y

x

T(h,k)

Q

R

p(x ,y )1 1 x

Q

R

y T(h,k)

o

∴The locus of (h,k) is xx1+yy

1 = a2, which is the equation of polar of P.

1 If the polar of P with respect to the circle x2+y2 = a2 touches the circle (x–f)2+(y–g)2 = b2 then itslocus is given by the equation.(a) (fx+gy–a2)2 = a2(x2+y2) (b) (fx+gy–a2)2 = b2(x2+y2)(c) (fx–gy–a2)2 = a2(x2+y2) (d) None of these.

2 The pole of the line 3x+4y = 45 with respect to the circle x2+y2–6x–8y+5 = 0 is(a) (6,8) (b) (6,–8)(c) (–6,8) (d) (–6,–8)

3 The pole of the chord of the circle x2+y2 = 16 which is bisected at the point (–2,3) with respectto the circle is

(a) ⎟

⎠

⎞

⎜

⎝

⎛

13

48,

13

32–(b) ⎟

⎠

⎞

⎜

⎝

⎛

13

48,

13

32

76

(c) ⎟

⎠

⎞

⎜

⎝

⎛

13

48–,

13

32–(d) None of these

4 The coordinates of the poles of thre common chord of the circles x2+y2 = 12 andx2+y2–5x+2y–2 = 0 with respect to the circle x2+y2 = 12 are

(a) ⎟

⎠

⎞

⎜

⎝

⎛

5

12–,6 (b) ⎟

⎠

⎞

⎜

⎝

⎛

5

12–,6–

(c) ⎟

⎠

⎞

⎜

⎝

⎛

5

12,6 (d) None of these

Q.5 The matching gridI The number of common tangents to the circles

x2+y2–6x–2y+9 = 0 and x2+y2–14x–8y+61 = 0 is (a) 3II The number of common tangents to the circles

x2+y2 = 4 and x2+y2–8x+12 = 0 is (b) 4III The number of common tangents to the circles

x2+y2 = 4 and x2+y2–6x–8y–24 = 0 is (c) 2IV The number of tangents to the circle

x2+y2–8x+6y+9 = 0 which pass through the point (3,–2) is (d) 1

Answer I →b(4) II →a(3)III →d(1) IV →c(2)

Q.6 If the line x cosθ +ysinθ = 2 is the equation of a transverse common tangent to the circles x2+y2

= 4 and x2+y2–6 3 x–6y+20 = 0, Then the value of θ is

(a) 5 π /6 (b) 2 π /3 (c) π /3 (d) π /6Q.7 Two circles of radii 4cm and 1cm touch each other externally and θ is the angle contained by their

direct common tangents. Then sinθ is equal to

(a)25

24(b)

25

12

(c)4

3(d) None of these.

Answer1. b 2. a 3. a 4. a 6. d 7.a

EXAMPLES1 If the tangent at the point P on the circle x2+y2+6x+6y = 2 meets the straight line

5x–2y+6 = 0 at a point Q on the y–axis, then the length of PQ is

(a) 4 (b) 2 5 (c) 5 (d) 3 5Solution CQ (0,3) is a point where the line 5x–2y+6 = 0 meets y–axis and the trangent drawn at PPQ is a tangent

77

(–3,–3)

Q(0, 3)

O

y

x

P

length of tangent PQ = 1S = 2–36090 ×+++

= 2–189 +

= 25= 5

2 If the two circles x2+y2+2gx+2fy = 0 and x2+y2+2g1x+2f

1y = 0 and touch each other, then

(a) f1g = fg

1(b) ff

1 = gg

1

(c) f2+g2 =f12+g

12 (d) None of these

Solution (a)

Centres of the circles are (–g,f) and (–g1,–f

1)and radius of the circles are 22 fg + and 2

12

1 fg +If two circles touch each other then|C

1C

2| = |r

1+r

2| (external)

or |r1–r

2| (internal)

∴ 21

21 )f–f()g–g( + = 22 fg + ± 2

12

1 fg +Squaring both sides

r1 r2

C1 C2

(–g,–f) (–g ,–f )11 or

(g–g1)2+(f–f

1)2 = g2+f2+g

12+f

12 ± 2 22 fg + 2

12

1 fg +

g2+g12–2gg

1+f2+f

12–2ff

1 = g2+f2+g

12+f

12 ± 2 2

122

122

122

12 ffgffggg +++

– (gg1+ff

1) = ± 2

122

122

122

12 ffgffggg +++

g2g12+f2f

12+2gg

1ff

1=g2g

12+g2f

12+f2g

12+f2f

12

g2f12+f2g

12–2gfg

1f

1 = 0 = (gf

1–fg

1)2

∴ gf1–fg

1 =0⇒gf

1=fg

1

78

3 The circles (x–a)2+(y–b)2 = c2 and (x–b)2+(y–a)2 = c2 touch each other, then

(a) a=b ± 2c (b) a = b ± c2(c) a = b ± c (d) None of theseSolutionDistance between centres = |r

1 ± r2|

Distance between (a,b) and (b,a) = |2c| or 0

22 )a–b()b–a( + = ± 2c

2 (a–b) = ± 2c

a–b = ± 2 c

a=b ± 2 c

79

CIRCLE - VII

Intersection of Two Circles

Angle of intersection of two circlesLet the two circles S ≡ x2+y2+2gx+2fy+c = 0 and S′ ≡ x2+y2+2g

1x+2f

1y+c

1=0 intersect

each other at the point P and Q. The angle θ between two circles S = 0 and S′ = 0 isdefined as the angle between the tangents to the two circles at the point of intersection.

θ must be taken acute angle .

C1 and C

2 are the centres of circles S = 0 and S′ = 0 then C

1(–g,–f) and C

2(–g

1,–f

1) and their

radii

r1 = c–fg 22 + & r

2=

1

21

21 c–fg +

C(-g,-f)

1

A

C(-g ,-f )

2

1 1

r1 r2

B

P

QA'B'

S=0S'=0

θ

Let d = |c1c

2| = 2

12

1 )f–f()g–g( + = 1

21

21

21

2 ff2–ffgg2–gg +++

C1P ⊥A A′ , C

2P⊥B B′ since radius is perpendicular to the tangent at the point of contact

ie. ∠ C1P A′ = 900 and ∠ C

2P B′ = 900

∴ ∠ C1P B′ = 900 - θ and ∠ C

2P A′ = 900– θ

Hence ∠ C1PC

2 = 900–θ +θ +900– θ = 1800– θ

Now in Δ C1PC

2

cos(1800– θ ) = 21

222

21

rr2

d–rr +(cosine rule)

cos θ = 21

222

21

rr2

d–rr +

If the angle between the circles is 900 ie.,θ = 900, then cos900 = 0 Then the circles are said to beorthogonal circles or the circles cut each other orthogonally.∴ r

12+r

22–d2 = 0

g2+f2–c+g12+f

12–c

1–g2–g

12+2gg

1–f2–f

12+2ff

1= 0

2gg1+2ff

1= c+c

1

It is a condition for two circles to be orthogonalExamples :1 Find the angle between the circles

80

S ≡x2+y2–4x+6y+11= 0 and S′ = x2+y2 –2x+8y+13 =0

SolutionHere S ≡x2+y2–4x+6y+11= 0 and S′ = x2+y2 –2x+8y+13 =0The centre of these circles are C

1(2,–3) and C

2(1,–4) respectively.

The radius of these circles are r1 = 11–94 + = 2

and r2 = 13–161 + = 2

Distance between the centres C1C

2 = d = 22 11 + = 2

Angle between two circle is cos θ = 21

221

22

21

rr2

)cc(–rr + =

222

2–42

××+

= 2

1=cos

4

π

∴ θ = 4

π

2 Find the equations of the two circles which intersect the circles x2+y2–6y+1 = 0 and x2+y2–4y+1 = 0 orthogonally and touch the line 3x+4y+5 = 0

SolutionLet the required equation of circle be x2+y2+2gx+2fy+c = 0This circle intersect orthogonally with circles x2+y2–6y+1 = 0 and x2+y2–4y+1 = 0Condition for orthogonality is 2gg

1+2ff

1 = c+c

1

∴ 0 + 2f(–3) = C+1 and 0+2f(–2) = c+1–6f = c+1 –4f=c+1

∴–6f = –4f ⇒ f= 0 and c = –1Equation of circle is x2+y2+2gx–1 = 0

Centre is (–g,0) and radius 1g2 +Since the line 3x+4y+5 = 0 touches the circle

∴ distance of this line from the centre must be equal to radius 1g 2 +

∴ 169

5g3–

++

= 1g 2 +

5–3g = 1g5 2 +squaring25+9g2–30g=25g2+2516g2+30g = 02g(8g+15) =0

g = 0 or g = 8

15–

Hence equations of circles are

81

x2+y2 –1 = 0 and x2+y2 – 4

15x–1 = 0

x2+y2–1 = 0 and 4x2+4y2–15x–4 = 03 Prove that the two circles, which pass through (0,a) and (0,–a) and touch the line y = mx+c, will

cut or thogonally if c2= a2(2+m2)Solution

Let the equation of the circles bex2+y2+2gx+2fy+d = 0This circle passes through the points (0,a) and (0,–a)∴a2+2fa+d= 0 __________(1) and a2–2fa+d = 0 _________(2)(1) (2)4fa = 0∴ f= 0 and d = –a2

∴The equation of circle is x2+y2+2gx–a2 = 0

Centre of this circle is (–g,0) and radies 22 ag +Since line y = mx+c touches the circle

∴1m

cmg–2 +

+ = 22 ag +

c–mg = 22 ag + 1m2 +Squaringc2+m2g2–2mcg = g2m2+g2+a2m2+a2

g2+2mcg+a2(1+m2)–c2 = 0It is a quadratic in g∴ product of the roots g

1g

2 = a2(1+m2)–c2

Sum of roots g1+g

2 = –2mc

Now the equations of the two circles represented are x2+y2+2g1x–a2=0 and x2+y2+2g

2x–a2=

0These two circles will be orthogonal if

2g1g

2 = – a2–a2

g1g

2= –a2

But g1g

2 = –c2+a2(1+m2)

∴–c2+a2(1+m2) = – a2

or c2 = a2 (2+m2)Which is the required condition

4 If the angle of intersection of the circles x2+y2+x+y = 0 and x2+y2+x–y = 0 is θ , then

equation of the line passing through (1,2) and making an angleθ with the y–axis isSolution

Let C1 and C

2 be the centres of given circles C

1⎟

⎠

⎞

⎜

⎝

⎛

2

1–,

2

1– and C

2 ⎟

⎠

⎞

⎜

⎝

⎛

2

1,

2

1–

82

Also radius these two circles are r1 =

4

1

4

1 + = 2

1

2

1 =

and r2 =

4

1

4

1 + = 2

1

cosθ = 21

222

21

rr2

d–rr +

=

2

1

2

12

1–2

1

2

1

×

+

= 0

∴ θ = 2

π

∴ Required line is parallel to x–axis and it passes through (1, 2)∴ Equation of line is y = 2.

5 The equation of a circle is x2+y2 = 4. Find the centre of the smallest circle touching the circle and

the line x+y = 25

yy

2

SolutionHere OA = 2 radius of circle x2+y2 = 4 with centre (0,0)

The distance of (0,0) from x+y = 25 is

2

25– = 5

∴The radius of the smallest circle = 2

2–5 =

2

3

and OC = 2+2

3 =

2

7

The slope of OA = 1 = tan θ

135

(-2,0) x+y= 5(2,0)

0

450

2

83

∴cosθ = 2

1 and sin θ =

2

1

∴ Centre (0+OC cos θ , O+OC sin θ ) = ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

22

7,

22

7

85

CIRCLE - VIIIRadical Axis

Radical axisThe radical axis of two circles is the locus of a point which moves such that the lengths of thetangents drawn from it to the two circles are equal.

P

T1T2

C1C1 C2

PT1=PT2

Radical axis

P

T1T2

TC1C1 C2

PT1=PT2

Radical axis

P

T1T2

C1C1 C2

B

A

PT1=PT2

Radical axis

Consider S = x2+y2+2gx+2fy+c = 0 and S′ = x2+y2+2g1x+2f

1y+c

1=0

Let P(x1, y

1) be any point such that |PT

1|=|PT

2|

or (PT1)2 = (PT

2)2

x12+y

12+2gx

1+2fy

1+c = x

12+y

12+2g

1x

1+2f

1y

1+c

1

2x1(g–g

1)+2y

1(f–f

1)+c–c

1 = 0 (1)

∴ locus of P(x1, y

1) is a straight line and equation (1) represents its equation.

Equation of radical axis is 2x(g–g1)+2y(f–f

1)+c–c

1 = 0

Properties of radical axis1 Radical axis is perpendicular to the line joining the centres of the given circles.

(Slope of radical axis) × slope of line joining centres = –1

)f–f(

)g–g(–

1

1×

1

1

g–g

f–f = – 1

2 The radical axis bisects common tangents of two circles

P

M

P’C D

BA

As M lies on radical axis and AB is a tangent to the circles ∴ AM = BM.Hence radical axis bisects the common tangents

3 If two circles cut a third circle orthogonally, then the radical axis of the two circles will passthrough the centre of the third circle, or the locus of the centre of a circle cutting two given circlesorthogonally is the radical axis of the given two circlesCP = CQ

86

P Q

C

C1 C2

S=0 0S =′

0S =′′

Hence C lies on the radical axis of the circles S = 0 and S′ = 0

4 The position of the radical axis of the two circles geometrically

radical axis common tansverse tangent

radical axis common chord

radical axis direct common tangent

Radical CentreThe radical axes of three circles, taken in pairs, meet in a point, which is called their radicalcentre.Let the three circles be S

1 = 0, S

2 = 0 S

3 = 0

OL, OM, ON be radical axes of the pair sets of circles{S1 = 0, S

2 = 0}, {S

2=0, S

3=0},

{S3=0, S

1=0} respectively

N

S =01

S =03

S =02

M

L

Equations of OL, OM and on are S1–S

2 = 0 S

2–S

3 = 0 and S

3–S

1 = 0

Family of lines passes through point of intersection of lines S1–S

2 = 0 and S

2–S

3 = 0 is

S1–S

2+ λ (S

2–S

3) = 0

If λ= 1 then the line becomes S3–S

1 = 0

O

87

∴The three lines are concurrent at OO is called radical centre .

Properties of radical centre1 Coordinates of radical centre can be found by solving the equations S

1=S

2 = S

3 = 0

2 The radical centre of three circles described on the sides of a triangle as diameters is the orthocenterof the triangle.

A

B C

A

CB

3 The radical centre of three given circles will be the centre of a fourth circle which cuts all the threecircles orthogonally. and the radius of the fourth circle is the length of tangent drawn from radicalcentre of the three given circles to any of these circles(x–y)2+(y–k)2 = r2 is the fourth circle with centre (h,k) and radius rCentre (h,k) is the radical centre of three circles and radius r is the length of tangent to one ofthese three circles from radical centre (h,k)

Example:1 The equation of the three circles are given x2+y2 = 1 ,x2+y2–8x+15 =0, x2+y2+10y+24 = 0

Determine the coordinates of the point P such that the tangents drawn from it to the circles areequal in lengthSolution : We know that the point from which lengths of tangents are equal in length is radicalcentre of the given three circlesRadical axis of first two circles isS

1–S

2 = 0

8x–16 = 0 or x–2 = 0___________________(1)and radical axis of second and third is–8x–10y–9 = 0_________________________(2)Solving (1) and (2) equations we getx=2, –16–10y–9 = 0

10y = –25

y = 10

25– =

2

5–

∴ Radical centre P is (2, –5/2)2 Find the radical centre of circles x2+y2+3x+2y+1 = 0,x2+y2–x+6y+5 = 0 and x2+y2+5x–8y+15

88

= 0. Also find the equation of the circle cutting them orthogonally.

Solution :Let S

1 ≡ x2+y2+3x+2y+1 = 0

S2 ≡ x2+y2–x+6y+5 = 0

S3 ≡ x2+y2+5x–8y+15 = 0

Equations of radical axisS

1–S

2 = 0 ⇒ 4x–4y–4 = 0 or x–y–1 = 0___________(1)

S2–S

3 = 0 ⇒ –6x+14y–10 = 0 or –3x+7y–5 = 0 ________(2)

Solve equations (1) & (2) we get (3, 2) as radical centre.(1) ×3–(2)×1

08–y4

––

03y3x3–

05–y7x3–

=+

=++=+

��

y = 2x–2–1= 0 ⇒ x = 3radius of fourth circle cutting these three circles orthogonally is length of tangent from this centreto any one circle

∴r = 1S

= 12.23.323 22 ++++

= 14949 ++++

= 27

= 3 3

∴Equation of circle is (x–3)2+(y–2)2= (3 3 )2

x2+y2–6x–4y–14 = 03 Find the radical centre of three circles described on the three sides 4x–7y+10 = 0, x+y–5= 0,

7x+4y–15 = 0 of a triangle as diameters.

Solution : We know that radical centre of 3 circles described on the three sides of a triangle asdiameters is orthocenter of the triangle.L

1 ≡ 4x–7y+10 = 0________ (1)L

2 ≡ x+y–5 = 0___________(2)L

3 ≡ 7x+4y–15 = 0__________(3)Slope of line L

1 is 4/7 and slope of line L

3 = –7/4 since the product of these slopes is –1

∴L1 and L

3 are perpendicular.

We know that orthocentre of a right angled triangle is the vertex where it has right angle.Solving (1) & (3) to get orthocentre(1)×4+(2)×7

89

065–x65

0105–y28x49

040y28–x16

==+

=+

x = 14–7y+10 = 07y+14y = 2∴Radical centre is (1, 2)

91

CIRCLE - IX

Co - Axial System of Circles

Co - Axial System of Circles

A system of circles or family of circles, every pair of which have the same radical axis arecalled co-axial circles

1 The equation of family of co-axial circles when the equation of radical axis and one circle aregiven

S=0S+ =0Lλ

S+ =0Lλ

L = ax+by+c = 0

S ≡ x2+y2+2gx+2fy+c=0

Then equation of co-axial circle is S+λL = 0

2 The equation of co-axial system of a circles where the equation of any two circles of the systemare

S =01

S+ =0Lλ

S + 1 ( - =0S S )1 2λ

S =02

S + 2 (S -S )=01 2λ

S1 = x2+y2+2gx+2fy+c = 0

S2 = x2+y2+2g

1x+2f

1y+c

1 = 0

respectively is S1+ λ (S

1–S

2) = 0

and S2+λ (S

1–S

2) = 0 ( λ ≠ – 1)

92

S1+λ S

2 = 0 ( λ ≠ – 1)

3 The equation of a system of co-axial circles in the simplest form is x2+y2+2gx+c = 0 where g isvariable and c is a constantThe common radical axis is the y-axis (centre on x-axis)The equation of system of co-axial circles in the simplest form is x2+y2+2fy+c = 0 where f is avariable and c is a constantThe common radical axis is the x–axis (centre on y axis)

x

y

o

y

xo

Examples :1 Find the equation of the system of circles co-axial with the circles x2+y2+4x+2y+1=0 and x2+y2–

2x+6y–6 = 0. Also find the equation of that particular circles whose centre lies on radical axis.Solution :

Given circles areS

1 ≡x2+y2+4x+2y+1=0S

2 ≡x2+y2–2x+6y–6 =0S

1–S

2 = 0

6x–4y+7 = 0System of co-axial circle is S

1+λ (S

1–S

2) = 0

x2+y2+4x+2y+1+ λ (6x–4y+7) = 0

x2+y2+2x(2+3 λ )+2y(1–2 λ )+1+7 λ = 0

Centre of this circle is (–(2+3λ ), – (1–2 λ ))lies on radical axis

∴6(–2–3 λ )+4(1–2 λ )+7 = 0

–12–18 λ+4–8λ +7 = 0

–1 –26 λ = 0

λ = 26

1–

∴Required particular member of co-axial circle is 26(x2+y2)+98x+56y+19 = 02 If the circumference of the circle x2+y2+8x+8y–b = 0 is bisected by the circle

93

x2+y2–2x+4y+a = 0 then a+b equal to(a) 50 (b) 56 (c) –56 (d) –34

Solution : (c)

Equation of radical axis (common chord of these circles) is10x+4y–b–a = 0Centre of first circle is (–4, –4)Since second circle bisects the first circleTherefore centre of first circle must lie on common chord.∴ 10(–4)+4(–4)–b–a= 0

–40–16–(a+b) = 0∴ a+b = –56

3 The equation of the circle passing through the point of intersection of the circles x2+y2–4x–2y = 8and x2+y2–2x–4y = 8 and the point (–1,4) is(a) x2+y2+4x+4y–8 = 0 (b) x2+y2–3x+4y+8 = 0(c) x2+y2+x+y–8 = 0 (d) x2+y2–3x–3y–8 = 0

Solution : (d)

Equation of any circle passing through the point of intersection of the circles isx2+y2–4x–2y–8+λ (x2+y2–2x–4y–8) = 0This circle passes through the point (–1,4

∴1+16+4–8–8+ λ (1+16+2–16–8) = 0

5–5 λ = 0

λ = 1Required circle is x2+y2–3x–3y–8 = 0

4 If the common chord of the circles x2+(y–b)2 = 16 and x2+y2 = 16 subtends a right angle at theorigin then b =

(a) 4 (b) 4 2 (c) –4 2 (d) 8

Solution :

The equation of common chord isS–S

1 = 0

(y–b)2–y2 = 0b2–2by = 0b(b–2y) = 0 b = 0 ,so

∴ b = 2y or 1 = b

y2

The combined equation of the straight lines joining the origin to the points if intersection of

y = b/2 and x2+y2 = 16 2

b

y2⎟

⎠

⎞

⎜

⎝

⎛

⇒b2x2+(b2–64)y2 = 0

This equation represents a pair of perpendicular lines

∴ b2+b2–64 = 0 ⇒ b = ± 4 2

95

CIRCLE - XLimiting Points

Limiting PointLimiting points of a system of co-axial circles are the centres of the point circles belonging tothe family (circles whose radii are zero are called point circle)

1 Limiting points of the co-axial systemLet the circle is x2+y2+2gx+c = 0Where g is a variable and c is constant

∴centre (–g,0) and radius c–g2 respectively

Let c–g2 = 0 radius

g2–c = 0g2 = c

g = c±

Thus we get the two limiting points of the given co-axial system as ( )0,c �

& ( )0,c– �

The limiting points are real and distinct, real and coincident or imaginary accordingas C>,=, <0.

2 System of co-axial circles whose limiting points are givenLet (α ,β ) and ( γ ,δ ) be the two given limiting pointsThen corresponding circles with zero radii are

(x– α )2+(y–β )2 = 0 = x2+y2–2 α x–2β y+ α 2+β 2 = 0

(x– γ )2+(y–δ )2 = 0 = x2+y2–2 γ x–2 δ y+ γ 2+δ 2 = 0System of co-axial circle equation is

x2+y2–2 α x–2β y+ α 2+β 2+λ (x2+y2–2 γ x+2 δ y+ γ 2+δ 2) = 0 ( λ ≠ –1)

centre of this circle is ⎟

⎠

⎞

⎜

⎝

⎛

λ+δλ+β

λ+γλ+α

1,

1 _____________(1)

and radius= ( )

λ+λδ+λγ+β+α

⎟

⎠

⎞

⎜

⎝

⎛

λ+δλ+β+⎟

⎠

⎞

⎜

⎝

⎛

λ+γλ+α

1–

11

222222

= 0

After solving find λ substitute in (1)We get the limiting point of co-axial system.

Properties of Limiting points1 The limiting point of a system of co-axial circles are conjugate points with respect to any member

of the system.Let the equation of any circle bex2+y2+2gx+c = 0________________(1)

limiting points of (1) are ( )0,c �

, ( )0,c– �

96

The polar of the point ( )0,c �

is

x c +g(x+ c )+c = 0

(x+ c ) (g+ c ) = 0

x+ c =0 ⇒x = – c

∴ ( )0,c– �

) lies on this.

Similarly ( )0,c lies on polar with respect to ( )0,c– �

∴ These limiting points ( )0,c– and ( )0,c are conjugate points

2 Every circle through the limiting points of a co-axial system is orthogonal to all circles of thesystemLet the equation of any circle x2+y2+2gx+c = 0 where g is a variable and c is a constant Limiting

point of this circle are ( )0,c– and ( )0,c

Now let, x2+y2+2g1x+2f

1y+c

1 = be any circle passing through the limiting points ( )0,c– and

( )0,c .

∴ c –2g1 c +c

1 = 0_______(3) and c+2g

1 c +c1 = 0________(4)

Solving (3) and (4). We get g1 = 0 and c

1 = –c

∴ The equation becomesx2+y2+2f

1y–c = 0

applying condition of orthogonality2gg

1+2ff

1=c+c

1

o + o = c – cHence condition is satisfied for all values of g

1 and f

1

Examples :

1. If the origin be one limiting point of a system of co-axial circles of which x2+y2+3x+4y+25=0 is amember, find the other limiting point.

Solution :Equation of circle with origin (0,0) as limiting point is x2+y2=0Given that one member of system of co-axial circle is x2+y2+3x+4y+25 = 0∴ The system of co-axial circles is

x2+y2+ x1

3

λ+ + y1

4

λ+ + λ+1

25 = 0

centre ( ) ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+−

+−

λλ 1

2,

12

3

radius ( )214

9

λ+ + ( )21

4

λ+ – ( )λ+1

25 = 0

97

( )214

25

λ+-

λ+1

25 = 0

1 – 4(1+λ ) = 0

1 + λ = 1/4

λ = 1/4 – 1 = –3/4

∴centre (–6,–8) is the other limiting point of the system .

Example 2

Find the radical axis of co-axial system of circles whose limiting points are (–1,2) and (2,3).

Solution :

Equations of circles with limiting points (–1,2) and (2,3) are(x+1)2+(y–2)2 = 0, x2+y2+2x–4y+5 = 0 ___________(1)(x–2)2+(y–3)2 = 0, x2+y2–4x–6y+13 = 0 ___________(2)respectivelyEquation of radical axis of (1) and (2) isS

1–S

2 = 0

6x+2y–8 = 03x+y–4 = 0

Example 3

Find the equation of the circle which passes through the origin and belongs to the co-axial ofcircles whose limiting points are (1,2) and (4,3)

Solution :

Equation of circles with limiting points (1,2) and (4,3) are(x–1)2+(y–2)2 = 0 ⇒ x2+y2–2x–4y+5 = 0(x–4)2+(y–3)2 = 0 ⇒ x2+y2–8x–6y+25 = 0System of co-axial of circles equation isx2+y2–2x–4y+5+ λ (x2+y2–8x–6y+25) = 0______________(1)equation (1) passes through origin

∴ 5+25λ = 0

∴ λ = –1/5Substituting in (1) we get4(x2+y2)–2x–14y = 02x2+2y2–x–7y = 0

99

CIRCLE - XIProblem Solving

Example 1Find the equation of the image of the circle x2+y2+16x–24y+183 = 0 by the line mirror4x+7y+13 = 0

Solution :The given circle and line arex2+y2+16x–24y+183 = 0_________(1) and 4x+7y+13 = 0_____________(2)Centre and radius if the circle are

(–8, 12) and 183–14464 + = 25 = 5 respectively.

Equation of line C1C

2 is 7x–4y+k = 0 it passes through (–8,12)

∴ –56–48+k = 0k = 104

Equation of line C1C

2 is 7x–4y+104 = 0__________(3)

To get the coordinates of M. Solve the equation(2) & (3)(2)×4+(3)×7

0780x65

0728y28–x49

052y28x16

=+=+

=++

��

x = –12put the value of x in (2) we get–48+7y+13 = 07y= 35y = 5∴ coordinate of M is (–12,5)M is the midpoint of C

1and C

2

∴–12 = 2

8–h⇒h = –16

5= 2

12k + ⇒ K = –2

∴Equation of imaged circle is(x+16)2+(y+2)2 = 25

100

x2+y2+32x+4y+235 = 02. The circle passing through the point (–1,0) and touching the y–axis at (0,2) also passes

through the point

(a) ⎟

⎠

⎞

⎜

⎝

⎛ 10,2

3–(b) ⎟

⎠

⎞

⎜

⎝

⎛ 12,2

5–(c) ⎟

⎠

⎞

⎜

⎝

⎛

25

,23–

(d) (–4,0)

Solution :Family of circles passing through a point (0,2) and touching line x= 0 (y–axis) is

(x–0)2+(y–2)2+ λx = 0It passes through (–1,0)

∴ 1+4– λ = 0

∴ λ = 5

∴equation of circle isx2+y2+5x–4y+4 = 0If also passes through A(–x

1,0)

∴x12–5x

1+4 = 0

(x1–4)(x

1–1)=0

x1= 4, x

1= 1

∴ it also passes through A (–4,0)

3. Two parallel chords of a circle of radius 2 are at a distance 13 + apart. If the chords subtend

at the centre, angles of k

π and

k

2π, where k>0 then the value of [k] is...............

( [k] denotes the greatest integer)Solution :

Let k2

π = α =

2

1∠ AOB = ∠ AOM

Then ∠ CON = 2 αIn Δ AOM

cos α = 2

x

In Δ CON

cos2 α = 2

x–13 +

cos2 α = 2cos2 θ –1

cos2 α = 2 4

x2

–1

∴2

x–13 + =

2

x2

–1

o(–x ,0)1 (–1,0)

A B

(0,2)

x

y

A B

CD

2

2

M

N

x

0

α α

2α 2αx–13+

–13+

101

x–13 + = x2 – 2

x2+x– 3–3 = 0

∴ x = ( )

2

33411– ++±

= 2

34131– +±

= ( )2

1321– +±(∴13+4 3 = ( )2132 + )

x = 2

1321– ++

x = 3

cos α = 2

3 = cos

6

π

α = 6

π

∴ Required angle = k

π = 2 α =

3

π k=6 thus [k] =6

4. Let ABC and AB C′be two non-congruent triangles with sides AB = 4, AC = A C′ = 22 and

angle β = 30°. The absolute value of the difference between the areas of these triangles isSolution :

Draw circle through AC C′ and AB intersect the circle and P

In Δ ABD =AB

AD = sin300

4

AD =

2

1

∴AD = 2 = DC = C′DDifference of areas of Δ ABC and Δ AB C′ is Δ AC C′

∴ar( Δ AC C′ ) = 2

1 ×4×2 = 4 sq.u.

4. The centres of two circles C1 and C

2 each of unit radius are at a distance of 6 units from each

other. Let P be the mid-point of the line segment joining the centres of C1 and C

2 and C be a

circle touching circles C1 and C

2 externally. If a common tangents to C

1 and C passing through P

is also a common tangent to C2 and C

1 then the radius of the circle C is

C'

D

B

2

A'

2

4

2

2222C30°

⇒

P

1

102

Solution :In Δ CPC

2

CP2 = (CC2)2–(C

2P)2

k2 = (r+1)2 –9k2=r2+2r–8_________________(1)In Δ PQC

2

PQ 2= 32–12

= 8∴ In Δ CPQ k2 = r2+8______________(2)From (1) & (2)r2+8 = r2+2r+2r–816 = r∴ r = 8

5. A straight line through the vertex P of a triangle PQR intersects the side QR at the point S and thecircumcircle of the triangle PQR at the point T. If S is not the centre of the circumcircle then

(a)PS

1+

ST

1 < SR.QS

2(b)

PS

1 +

ST

1> SR.QS

2

(c)PS

1+

ST

1 < QR

4(d)

PS

1+

ST

1> QR

4

Solution :Points P, Q, T, R are concyclic∴PS.ST = QS.SR

⇒ 2

STPS +≥ ST.PS (AM ≥ GM)

∴PT ≥ 2 ST.PS

and PS

1+

ST

1 ≥ ST.PS

2 = SR.QS

2(Dividing by PS.ST)

Also, 2

SRSQ +≥ SR.SQ (AM≥GM)

⇒ 2

QR≥ SR.SQ

⇒ SR.SQ

1≥ QR

2

⇒ SR.SQ

2≥ QR

4

C

r

k

C1

P

Q

C21

3

8

6

P

R

Q 3

T

S

103

PS

1+

ST

1 ≥ SR.QS

2≥ QR

4

6 Let ABCD be a quadrilateral with area 18, with side AB parallel to the side CD andAB = 2CD. Let AD be perpendicular to AB and CD. If a circle is drawn inside the quadrilateralABCD touching all the sides then its radius is(a) 3 (b) 2 (c) 3/2 (d) 1.

Solution :ABCD is a trapezium ( AB parallel to CD)

∴ar(ABCD) = 2

1×h (sum of parallel sides)

= 2

1×2r(2a+a)

18 = r×3aar = 6CB is a tangent to the circle∴equation of tangent is

y = a

r2–(x–2a)⇒2rx+ay–4ar = 0

It is a tangent to the circle (x–r)2+(y–r)2 = r2

∴ r = 22

2

ar4

ar4–arr2

+

+

r 22 ar4 + = 2r2–3ar

22 ar4 + = 2r–3a

Squaring4r2+a2=4r2+9a2–12ar12r = 8a ⇒ 3r = 2aar = 6

r = 3

29

3

a2 2

= 6

a2 = 9a = ± 3∴ r = 2

7 The radius of the least circle passing through the point (8,4) and cutting the circlex2+y2 = 40 orthogonally is

(a) 5 (b) 7 (c) 2 5 (d) 3 5

Y

R

OA B(2a,0)

C (a,2r)D(o,2r)

r

(r,r)

104

Solution :Let the circle be x2+y2+2gx+2fy+c = 0___________(1)Given circle is x2+y2 = 40 _____________________(2)These two circles are orthogonal∴c–40 = 0 ⇒ c = 4 0(1) passes through (8,4)64+16+16g+8f+40 = 0120+16g+8f = 0f+2g+15 = 0 or f = –(2g+15)

radius = c–fg 22 +

= 40–)15g2(g 22 ++For least circle radius must be minimumLet f(g) = g2+(2g+15)2 – 40 is minimum

f ′ (g) = 2g+4(2g+15) = 010g = –60g = –6

f ′′ (g) = 10>0 minimumf = –(–12+15) = –3Equation of circle in x2+y2–12x–6y+40 = 0

radius = 40–936 +

= 58 P is a point (a,b) in the first quadrant. If the two circles which pass throngh P and touch both the

co-ordinate axes cut at right angles, then(a) a2–6ab+b2 = 0 (b) a2+2ab–b2 = 0(c) a2–4ab+b2 = 0 (d) a2–8ab+b2 =0

Solution :Equation of the two circles be(x–r)2+(y–r)2 = r2

⇒x2+y2–2xr–2yr+r2 =0These two circles passes through (a,b)∴ (a–r)2+(b–r)2 = r2

a2+r2–2ar +b2+r2–2br–r2 = 0r2–2r(a+b)+(a2+b2) = 0It is a quadratic equation in r∴ r

1+r

2 = 2(a+b)and r

1r

2 = a2+b2

Condition for orthogonality is2g

1g

2+2f

1f

2 = C

1+C

2

2r1r

2+2r

1r

2=r

12+r

22

4r1r

2 = r

12+r

22

6r1r

2 = r

12+r

22+2r

1r

2

6r1r

2=(r

1+r

2)2

6(a2+b2) = 4 (a+b)2

105

6a2+6b2= 4a2+4b2+8ab2a2+2b2–8ab = 0a2+b2–4ab = 0

9 A circle S ≡ 0 passes throngh the common points of family of circles x2+y2+dx–4y+3 = 0(λ ∈R)and have minimum area then(a) area of S ≡0 is π squ.

(b) radius of director circle of S ≡0 is 2(c) Radius of director circle of S ≡ 0 for x–axis is 1 unit(d) S ≡ 0 never cuts |2x| = 1

Solution :(x2+y2–4y+3)+λ x = 0

∴x= 0 and y2–4y+3 = 0(y–3)(y–1) = 0y = 3 , 1∴ (0,3) (0,1) are common points.x2+y2+2gx+2fy+c = 0passion through (0,3) & (0,1)

0cf69 =++ ___________________(1)

0cf21 =++ ___________________(2)(1) – (2) we get

0f48 =+3c2–f == ��

∴x2+y2+2gx–4y+3 = 0

radius = 3–4g2 + = 1g2 +for minimum area radius must be minimumSince g2+1 is positive so g must be zero∴ radius = 1Area = πr2 = πsq.u.

Radius of director circle is 2 times the radius of the given circle.

∴Radius of director circle is 210 Area of part of circle x2+y2–4x–6y+12 = 0 above the line 4x+7y–29 = 0 is Δ ,

then [ Δ ] = ________ [.] is greatest integer function.Solution

Since line 4x+7y–29 =0passes through the centre (2,3) of the circle

∴ The line is a diameter of a circle with radius 12–94 + = 1

∴ area of semi circle is = 2

1πr2

= 2

1π

(2,3)

4x+7y-29=0

106

= 2

14.3=1.57

Hence[ Δ ] = [1.57] = 1

PRACTICE QUESTIONS

1. Let 2π

α0 << be a fixed angle. If P = ( )θsin,θcos and Q = ( ( )θ–αcos , ( )θ–αsin Q is obtained

from P by

a. clockwise rotation around origin through an angleαb. anti-clockwise rotation around origin through an angleαc. reflection in the line through origin with slope tanα

d. reflection in the line through origin with slope tanα /2

2. If the tangent at the point P on the circle x2+y2+6x+6y = 2 meets the straight line 5x–2y+6 = 0at a point Q on the y-axis, then the length of PQ is

a. 4 b.

52

c. 5 d. 533. The equations to the sides AB, BC, CA of a ABCΔ are drawn on AB, BC, CA as diameters.

The point of concurrence of the common chord isa. centroid of the triangle b. orthocenterc. circumcentre d. incentre

4. The number of rational point(s) (a point (a, b) is rational, if a and b both are rational numbers)

on the circumference of a circle having centre ( )e,π is

a. at most one b. at least two c. exactly two d. infinite5. The locus of a point such that the tangents drawn from it to the circle x2+y2–6x–8y = 0 are

perpendicular to each other isa. x2+y2–6x–8y–25 = 0 b. x2+y2+6x–8y–5 = 0c. x2+y2–6x+8y–5 = 0 d. x2+y2–6x–8y+25 = 0

6. If the two circles x2+y2+2gx+2fy = 0 and x2+y2+2g1x+2f

1y = 0 touch each other, then

a. f1g = fg

1b. ff

1 = gg

1

c. f2+g2 = 21

21 gf + d. none of these

7. The number of integral values ofλ

for which x2+y2+λ

x+(1–λ

)y+5 = 0 is the equation of acircle whose radius cannot exceed 5, isa. 14 b. 18 c. 16 d. none of these

8. The circle x2+y2+4x–7y+12 = 0 cuts an intercept on y-axis of lengtha. 3 b. 4 c. 7 d. 1

9. One of the diameter of the circle x2+y2–12x+4y+6 = 0 is given bya. x+y = 0 b. x+3y = 0 c. x = y d. 3x+2y = 0

10. The coordinates of the middle point of the chord cut off by 2x–5y+18 = 0 by the circlex2+y2–x+y2–54 = 0 are

a. (1, 4) b. (2, 4) c. (4, 1) d. (1, 1)

107

11. A variable chord is drawn through the origin to the circle x2+y2–2ax = 0. The locus of thecentre of the circle drawn on this chord as diameter isa. x2+y2+ax = 0 b. x2+y2+ay = 0c. x2+y2–ax = 0 d. x2+y2–ay = 0

12. If O is the origin and OP, OQ are distinct tangents to the circle x2+y2+2gx+2fy+c = 0, thecircumcentre of the triangle OPQ isa. (–g, –f) b. (g, f) c. (–f, –g) d. none of these

13. Equation of the normal to the circle x2+y2–4x+4y–17 = 0 which passes through (1, 1) isa. 3x+2y–5 = 0 b. 3x+y–4 = 0c. 3x+2y–2 = 0 d. 3x–y–8 = 0

14. The equation of the circle touching the lines xy = at a distance 2 unit from the origin is

a. x2+y2–4x+2 = 0 b. x2+y2+4x–2 = 0c. x2+y2+4x+2 = 0 d. none of these

15. The shortest distance from the point (2, –7) to the circle x2+y2–14x–10y–151 = 0 isa. 1 b. 2 c. 3 d. 4

16. The equation of the image of the circle (x–3)2+(y–2)2 = 1 by the mirror x+y = 19 isa. (x–14)2+(y–13)2 = 1 b. (x–15)2+(y–14)2 = 1c. (x–16)2+(y–15)2 = 1 d. (x–17)2+(y–16)2 = 1

17. If P and Q are two points on the circle x2+y2–4x–4y–1 = 0 which are farthest and nearestrespectively from the point (6, 5), then

a.

= 3,

5

22–P b.

=

5

19,

5

22Q

c.

=

5

11,–

3

14P d.

= 4,–

3

14–Q

18. A circle of the coaxial system with limiting points (0, 0) and (1, 0) isa. x2+y2–2x = 0 b. x2+y2–6x+3 = 0c. x2+y2 = 1 d. x2+y2–2x+1 = 0

19. If a variable circle touches externally two given circles, then the locus of the centre of thevariable circle isa. a straight line b. a parabolac. an ellipse d. a hyperbola

PASSAGE – 1

Let A ( )0,a≡ and B be two fixed points and P moves on a plane such

that PA = nPBOn the basis of above information, answer the following questions:

20. If 1n ≠ , then the locus of a point P is

a. a straight line b. a circle c. a parabola d. an ellipse21. If n = 1, then the locus of a point P is

a. a straight line c. a circle c. a parabola d. a hyperbola

108

22. If 0 < n < 1, thena. A lies inside the circle and B lies outside the circleb. A lies outside the circle and B lies inside the circlec. both A and B lies on the circled. both A and B lies inside the circle

23. If n > 1, thena. A lies outside the circle and B leis inside the circleb. A lies outside the circle and B leis inside the circlec. both A and B lies on the circled. both A and B lies inside the circle

24. If focus of P is a circle, then the circlea. passes through A and Bb. never passes through A and Bc. passes through A but does not pass through Bd. passes through B but does not pass through A

PASSAGE – 2For each natural number k, let Ck denotes the circle with radius k units and centre at the origin.On the Ck-, a particle moves k units in the counter clockwise direction. After completing itsmotion on Ck, the particle moves to C

kt1in some well defined manner, where k>0.The motion

of the particle continues in this manner.On the basis of above information, answer the following questions:

25. Let, K=1 the particle starts at (1, 0). If the particle crossing the positive direction of the x-axisfor the first time on the circle Cn, then n is equal toa. 3 b. 5 c. 7 d. 8

26. If Nk ∈ and

, the particle starts (–1, 0) the particle cross x-axis again ata. (3,0) b. (1,0) c. (4,0) d. (2,0)

27. If and

, the particle moves in the radial direction from circle Ck to CK+1. If particlestarts form the point (–1, 0), thena. it will cross the +ve y-axis at (0, 4)b. it will cross the –ve y-axis at (0, –4)c. it will cross the +ve y-axis at (0, 5)d. it will cross the –ve y-axis at (0, –5)

28. If and

, particle moves tangentially form the circle Ck to Ck+1, such that the lengthof tangent is equal to k units itself. If particle starts form the point (1, 0), thena. the particle will cross x-axis again at x = 3b. the particle will cross x-axis again at x = 4

c. the particle will cross +ve x-axis again at x =

d. the particle will cross +ve x-axis again at ( )4,22x ∈29. Let the particle starts from the point (2, 0) and moves 2/π units, on circle C2 in the counter-

clockwise direction, then moves on the circle C3 along the tangential path, let this straight line(tangential path traced by particle) intersect the circle C3 at the points A and B tangents drawn

109

at A and B intersect at

a.

22

9;

22

9b. ( )29,29 c. (9, 9) d. ( )2,2

Match Type:30. Observe the following columns:

Columns I Columns IIa. If the shortest and largest distance from the point p. M+L = 10

(10, 7) to the circle x2+y2–4x–2y–20 = 0 are Land M respectively, then

b. If the shortest and largest distance from the point Q. M + L = 20(3, –6) to the circle x2+y2–16x–12y–125 = 0 are Land M respectively, then

c. If the shortest and largest distance from the point r. M + L = 30(6, –6) to the circle x2+y2–4x+6y–20 = 0 are L s. M – L = 10and M respectively, then t. M – L = 26

31. Observe the following columns:Columns I Columns II

a. If the straight lines y = a1x+b and p. 4aa 2

221 =+

y = a2x+b ( )21 aa ≠ meet the

coordinate axes in concyclic points, thenb. If the chord of contact of the tangents drawn q. a1+a2 = 3

to x2+y2= b2 from any point on x2+y2 = ,

touches the circle

x2+y2 = ( )2122 aaa ≠ , then

c. If the circles x2+y2+2a1x+b = 0 and r. a

1a

2 = b

x2+y2+2a2x+b = 0 ( )21 aa ≠ and cuts s. a

1a

2 = 1

orthogonally, then t. a1a

2 = b2

ANSWERS

1. d 2. c 3. b 4. a 5. a 6. a7. c 8. d 9. b 10. a 11. c 12. d 13. b14. a 15. b 16. d 17. b 18. d 19. d 20. b21. a 22. a 23. b 24. b 25. c 26. c 27. c28. d 29. a30. a q,s b→ r,t c→ s31. a → p,q,s b → p,q,s,t c → p,q,r,s

111

PARABOLA - I

113

=Δ⇐

aa h gh b fg t c

114

PRACTICE QUESTIONS

117

PARABOLA - IIPARABOLA

A parabola is the locus of a point, whose distance from a fixed point is equal to the perpendiculardistance from a fixed straight line.

Let S be the focus, ZZ ′ be the directrix.Consider S (a, 0) and equation of ZZ ′ is x+a = 0.Axis of parabola is x-axis.Now according to definition.PS = PM

( )1

axya–x 22 +=+

( ) ( )222 axya–x +=+

ax4y2 = Vertex (0, 0)Tangent of latus rectum x = 0Extremities of latus rectum (a, 2a) , (a, –2a)Length of latus rectum.=4aFocal distance (SP) SP = PM = x + aParametric form x = at2, y = 2at, t is parameter.Focal distance – the distance of a point on the parabola from the focus.Focal chord – A chord of the parabola, which passes through the focus.Double ordinate – A chord of the parabola perpendicular to the axis of the parabola.Latus Rectum – A double ordinate passing through the focus or a focal chord perpendicular tothe axis of parabola.• Perpendicular distance from focus on directrix = half the latus rectum.• Vertex is middle point of the focus and the point of intersection of directrix and axis.• Two parabolas are said to be equal if they have the same latus rectum.

Other Standard Forms of Parabola:

S

P

A

Equation of curve: y2 = – 4ax x2= 4ay x2 = –4ayVertex (0, 0) (0, 0) (0, 0)

S (a,0)

118

Focus (–a, 0) (0, a) (0, –a)Directrix x – a = 0 y + a = 0 y – a = 0Equation of axis y = 0 x = 0 x = 0Tangent of vertex x = 0 y = 0 y = 0Parametric form (–at2, 2at) (2at, at2) (2at, –at2)

Position of a point with respet to Parabolay2 = 4ax y2 = – 4ax x2 = 4ay x2 = – 4ay

P (x ,y )11

S1 = y

12 – 4ax

1S

1 = y

12 + 4ax

1S

1 = x

12 – 4ay

1S

1 = x

12 + 4ay

1

S1< 0 → Inside S

1< 0 → Inside S

1< 0 → Inside S

1< 0 ← inside

S1> 0 → Outside S

1> 0 → Outside S

1> 0 → Outside S

1> 0 → Outside

S1= 0 → on the parabola S

1= 0 → on the parabola S

1= 0 → on the parabola S

1= 0 → on the parabola

Equation of Parabola when vertex is shifted.

I. Axis is Parallel to x-axis:Let vertex A be (p, q) then equation of parabola be(y – q)2 = 4a (x – p).

II. Axis is parallel to y-axis:Let vertex A be (p, q) then equation of parabola is(x – p)2 = 4a (y –q)

Example: 1 The equation of parabola is y = ax2 + bx + c, find its vertex, focus, directrix,Equation of a is, length of latus rectum.

y = a ca4

b–

a4

bx

a

bx 2

2

2

22 +

++

y – c = a

+ 2

22

a4

b–

a2

bx

(p,q)S

y

A

xO

z

z'

(p,q)

S

y

A

xOz z'

119

+=

+ c–

a4

by

a

1

a2

bx

22

X2 = 4 AY where X = a

1A4,

a4

ac4–byY,

a2

bx

2

=+=+ �

Vertex : X = 0, Y = 0 i.e.

a4

ac4–b–,

a2

b–

2

��

Focus: X = 0, Y = A i.e.

+

a4

b–ac41,

a2

b–

2

�

Equation of directrix: 0a4

1ac4–by

2

=++

Equation of axis: 0a2

bx =+

Length of latus rectum = a

1

Example: 2 The equation of parabola is y2 = ax + ay. Find its vertex, focus, directrix, axisand length of latus rectum.y2 – ay = ax

y2 – ay +4

a 2

= ax +4

a 2

+=

4

axa

2

a–y

2

Y2 = 4AX

Where: Y = y – 2

a, X = x +

4

a, 4A = a ie. A =

4

a

Vertex

2

a,

4

a–

Focus

2

a,0

Directrix 02

ax =+

120

Axis 02

a–y =

Length of latus rectum = a

PRACTICE QUESTIONS

1. The equation of parabola whose focus is at (–1,–2) and directrix is x–2y+3 = 0 isa. 4x2–y2–4xy+4x–32y–16 = 0 b. x2+4y2+4xy+x+6y+16 = 0c. 4x2+y2+4xy+4x+32y+16 = 0 d. 4x2+y2–4xy+4x–32y–1 = 0

2. The equation of parabola whose vertex is at (4, –1) and focus is (4, –3) isa. y2–8x+8y+24 = 0 b. x2–8x+8y+24 = 0c. y2–8x–8y+24 = 0 d. x2+8x–8y–24 = 0

3. The focal distance of a point on the parabola y2=8x is 8, then coordinates of the point (s) is/are

a. ( )6,34 b. ( )34,6 c. ( )6,–34 d. ( )34,–64. The equation of the parabola whose focus is (0, 0) and tangent at the vertex is x–y+1 = 0 is

a. x2+y2+2xy–4x+4y–4 = 0 b. x2+y2+4xy+4x+4y+4 = 0c. x2+y2–4xy+4x+4y–4 = 0 d. x2+y2–4xy–4x–4y–4 = 0

ANSWERS

1. c 2. b 3. b, d 4. a

121

PARABOLA - IIIPARAMETRIC FORM:

y2 = 4ax y2 = – 4ax x2 = 4ay x2 = – 4ayx = at2 x = – at2 x = 2at x = 2aty = 2at y = 2at y = at2 y = – at2

Parametric (at2, 2at) (– at2, 2at) (2at, at2 ) (2at, – at2 )coordinates

Properties of Focal chord:1. If the chord joining P(t

1) and Q(t

2) is the focal chord then t

1.t

2 = – 1.

P, S and Q lies on the focal chord∴ P, S and Q are collinear slope of PS = slope of SQ

a–at

at2

a–at

at222

221

1 =

1–t

t2

1–t

t222

221

1 =

22121

221 t–ttt–tt =

2t – 1t = 1t 2t ( 1t – 2t )

1221 t

1–tortt1– == ��

Extremities of a focal chord are (at2, 2at) and

t

a2–,

t

a2 .

2. Length of focal chord is2

t

1ta

+

PQ = PS + SQ

= at2 + a + 2t

a+ a

= a

++ 2

t

1t 2

2

= a2

t

1t

+

3. The length of the focal chord which makes an angleθwith the positive direction of x-axis is

4a cosec2θ .

We know PQ = a2

t

1t

+

(at ,2 )12 at1

(at ,2 )2

2 at2

S

y

P

Q

xO(a,o)

(at,2)2 at

S

y

P

Q

xO

−

ta2

,ta2

(a,o)

122

slope =

22

ta

–at

ta2

at2θtan

+=

slope = tanθ =

t1

–t

2

2cotθ= t

1–t

∴ PQ= a2

t

1t

+

= a

+

4t

1–t

2

= a [4 cot2θ+ 4]

= 4a cosec2θ ( )1θeccos ≥

• Minimum length of PQ = 4a ( )rectumlatus.e.i ��

4. Semi latus rectum of a parabola is the harmonic mean between the segments of any focal chordof the parabola.

SP = a + at2, SQ = a + 2ta

aat

t

ata

1

ta

a

1

ata

1

SQ

1

SP

12

2

2

2

2 ++

+=

++

+=+ =

a

1

SQSP

SQxSPx2a2

+=∴ ��

Semi latus rectum = Harmonic Mean of SP and SQ.

5. Circle described on the focal length as diameter touches thetangent at vertex.Equation of circle PS as diameter is(x – at2) (x – a) + (y – 2at)y = 0Equation of y-axis is x = 0After solving y2 – 2aty + a2t2 = 0

(y – at)2 = 0∴circle touches the y-axis at (0, at).

(at ,2 )2 at

S θ

y

P

Q

xO

−

ta2

,ta2

(at ,2 )2 at

S

y

P

Q

xO

−

ta2

,ta2

y

x(a,0)

123

(a,0)S

y

P

Q

MbxO

Example: 1 The length of a focal chord of the parabola y2 = 4ax at a distance b from thevertex is c, then

a. b2 = 4ac b. b2c = a3 c. b2c = 4a3 d. 4b2c = a3

Solution: PQ = 4a cosec2θ = c

In OMSΔ , sinθ=a

b

cosec2θ = 2

2

b

a

∴ c = 4a. 2

2

b

a b2c = 4a3

Ans: c

Example: 2 The coordinates of the ends of a focal chord of a parabola y2 = 4ax are (x1, y

1)

and (x2, y

2) then value of x

1 x

2 + y

1 y

2 is equal to

a. 3a2 b. – 3a2 c. a2 d. – a2

Solution: Let ( ) ( )11121 y,xat2,at ≡ and ( ) ( )222

22 y,xat2,at ≡ such that t

1t2 = –1

∴x1x

2 + y

1y

2 = 222

2122

221

2 a3–a4–atta4tta ==+Ans: b

PRACTICE QUESTIONS

1. The focus of the parabola x2+8x+12y+4 = 0 isa. (4, 2) b. (–2, –4) c. (2, 4) d. (–4, –2)

2. The equation of the parabola with vertex at (3, 2) and focus at (5, 2) isa. x2–8x–4y–28 = 0 b. y2–8x–4y–28 = 0c. x2+8x–4y–28 = 0 d. y2+8x+4y–28 = 0

3. The equation of the latus rectum of the parabola x2+4x+2y = 0 isa. 2y–3 = 0 b. 3y–2 = 0c. 2y+3 = 0 d. 3y+2 = 0

4. The equation of the parabola whose axis is parallel to x-axis and which passes through the points(0, 4), (1, 9) and (–2, 6) is

a. y2+5x–25y+139 = 0 b. 3y2+5x–25y+52 = 0c. 2y2–5x–25y+68 = 0 d. none of these

5. The parametric equation x = at2+bt+c, ctbtay 2 ′+′+′= represents

a. a circle b. a parabola c. an ellipse d. none of these

ANSWERS1. d 2. b 3. a 4. c 5. b

θ

125

PARABOLA - IVLine and Parabola:

y

xO

y

xO

y

xO

Let equation of line be y = mx + c and equation of parabola be y2 = 4ax.(mx + c)2 = 4axm2x2 + 2x(mc – 2a) + c2 = 0

D = ( ){ } 222 c.m.4–a2–mc2 �

= 4(4a2 – 4amc)If D < 0, line do not intersect parabola.i.e. a < mcIf D = 0 i.e. a = mc, line touches the parabola (condition of tangency)If D > 0 i.e. a > mc, line intersect the parabola at two points.

Equation of tangent (Point form)Equation of parabola y2 = 4axDifferentiate w.r.t.x

a4dx

dyy2 =

slope of tangent = 1y

a2

Equation of tangent y – y1 =

1y

a2 (x – x

1)

y y1 –

121 ax2–ax2y =

y y1 = 2ax + 2ax

1

y y1 = 2a (x+x

1)

Equation of tangent (Paramatric form)y.2at = 2a (x + at2)ty = x + at2

(x ,y )11

y

xO

(at ,2at)2

y

xO

126

Equation of tangent (slope form)

m

amxy +=

Point of contact

m

a2,

m

a2

• Equation of tangent to the parabola (y – k)2 = 4a (x – h) is

y – k = m(x – h) +m

a

Equation of tangent:Parabola Point form Pt. of contact Parametric form Pt. of contact slope form Pt. of contact

y2=4ax yy1=2a(x + x

1) (x

1, y

1) ty = x + at2 (at2, 2at) y = mx +

m

a

m

a2,

m

a2

y2= –4ax yy1=–2a(x + x

1) (x

1, y

1) ty = – x+ at2 (– at2, 2at) y = mx –

m

a

m

a2–,

m

a– 2

x2= 4ay xx1= 2a(y + y

1) (x

1, y

1) tx = y + at2 (2at, at2) x = my +

m

a

2m

a,

m

a2

x2= – 4ay xx1=–2a(y + y

1) (x

1, y

1) tx = – y + at2 (2at, – at2) x = my –

m

a

2m

a–,

m

a2–

Pair of Tangents from point (x1, y

1)

Let eqn of parabola be y2 = 4axS ≡ y2 – 4ax

S1≡ 2

1y – 4ax1

T ≡ yy1 – 2a(x+x

1)

Equation of pair of tangents is SS1 = T2 i.e.

(y2 – 4ax) ( )121 ax4–y = ( ){ }2

11 xxa2–yy +

Properties of Tangents:1 . Point of intersection of two tangents of the parabola:-

Equation of tangent at P is t1y = x + 2

1at

Equation of tangent at Q is t2y = x + 2

2atSolving these equations, we getx = at

1t

2, y = at

1 + at

2

A(at1t

2, a(t

1 + t

2))

2. Locus of foot of prependicular from focus upon any tangent is tangent at vertex:-Equation of tangent at P is ty = x + at2

P(x , y )1 1

A

B

A

P(t )1

Q(t )2

127

Let the tangent meet y-axis at Q then Q(0, at)

∴ slope of QS = t–a

at–=

slope of tangent = t

1

t

1x (– t) = –1 SQ ⊥ tangent

3. Length of tangent between the pt. of contact and the pointwhere tangent meets the directrix subtends right angle at focus:-Eqation of tangent at P(t)

ty = x + at2

Point of intersection with directrix x = – a is

t

a–at,a–

slope SP = 1–t

t2

a–at

at222 =

slope QS = t2–

1–t

a2–ta

–at 2

=

m1 m

2 = –1

PS ⊥ QS.4. Tangent at extremities of focal chord are perpendicular and intersect on directrix

(Locus of intersection point of tangents at extremities of focal chord is directrix)

Let P(at2, 2at) and

t

a2–,

t

aQ 2

Equation of tangent at P ty = x+at2 ......(1)

Equation of tangent at Q 2ta

xyt1

– +=

t

a–tx–y = ......(2)

Point of intersection of both tangents, we get after sloving (1) & (2) i.e.x+a = 0

A point lies on the directrix.

y

x

P(at ,2at)2

S(a,0)

Q

128

PRACTICE QUESTIONS

1. If the tangents to the parabola y2

= 4ax at the points (x1

, y1

) and (x2

, y2

) meet at the point (x3

, y3

)

then

a. 2123 yyy = b. 213 yyy2 += c.

213 y1

y1

y2 +=

d. none of these

2. A right angled triangle ABC is inscribed in parabola y2 = 4x, where A is vertex of parabola and

BAC∠ = 900. If AB =

5

, then area of ABCΔ is

a. 40 b. 10 c. 20 d. 54

3. The locus of the point ( )2k3,h3 + if it lies on the line x–y–1 = 0 is a

a. circle b. parabola c. straight line d. none of these4. The length of the chord of the parabola y2 = x which is bisected at the point (2, 1) is

a. 5 b. 4 c. 52 d. 255. If y = mx+c touches the parabola y2 = 4a (x+a), then

a.ma

c = b.ma

ac += c.ma2

c = d.ma

amc +=

ANSWERS1. a 2. c 3. b 4. c 5. d

129

PARABOLA - V

Example: 1 The focal chord to y2 = 16x is tangent to (x–6)2 +y2 = 2, then the possible value ofthe slope of this chord, are

a. {–1, 1} b. {–2, 2} c.

2

1,2– d.

2

1,–2

Solution: The focus of parabola is (4, 0). Let slope of focal chord be m. Equation of focalchord is y = m(x–4). It is tangent to the circle then

21m

m4–m62

=+

4m2 = 2(m2+1)2m2 = 2

1m ±=Ans: a

Example: 2 The curve represented by 1byax =+ , where a, b > 0 is

a. a circle b. a parabola c. an ellipse d. a hyperbola

by–1ax =

Solution: ax = 1+by by2–

(ax–by–1)2 = ( )2by2–

a2x2+b2y2+1–2abxy–2ax+2by = 4bya2x2–2abxy+b2y2–2ax–2by+1 = 0

1b–a–

b–bab–

a–ab–a

Δ

��

��

��

�

�

=

= a2(b2–b2)+ab(–ab–ab) –a(ab2+ab2)

= 0 – 2a2b2–2a2b2

0ba4– 22 ≠= �

h2–ab = (–ab)2 – (a2) (b2) = 0∴ It is a parabolaAns: b

130

Example: 3 The equation of the directrix of the parabola y2+4x+4y+2 = 0 is

a. x = –1 b. x = 1 c. x = 2

3–d. x =

2

3

Solution: y2 + 4y = –4x–2

(y+2)2 = –4

2

1–x

y2 = –4AX

Equation of directrix is X = A i.e. 12

1–x =

2x–3 = 0 or 2

3x =

Ans: d

Example: 4 The locus of the midpoint of the segment joining the focus to a moving point on theparabola y2 = 4ax is another parabola with directrixa. y = 0 b. x = –a c. x = 0 d. none of these

Solution: Let P(at2, 2at) lies on the parabolay2 = 4axMid point of PS is Q.

h = 2

at20k,

2

aat2 +=+

ta

k,t

a

a–h2 2 ==

2

a

k

a

a–h2

=

a(2h – a) = k2

Locus of (h, k) is y2 = 2a

2

a–x

Equaiton of directrix is2

a–

2

a–x =

x = 0Ans: c

131

Example: 5 The angle between the tangents drawn from the point (1,4) to the parabola y2 = 4xis

a.6

π

b.4

π

c.3

π

d.2

π

Solution: Equation of tangent of the parabola y2 = 4x is

m

1mxy +=

This equation passes through (1, 4) i.e.

m

1m4 +=

m2 – 4m+1 = 0m

1.m

2 = 1 and m

1 + m

2 = 4

Angle between the two tangents is21

21

mm1

m–mθtan

+=

( )3

2

12

11

4–4θtan

2

==+

=

3

π

θ =∴

Ans: cExercise: 6 Tangent to the curve y = x2+6 at a point (1, 7) touches the circle x2+y2+16x+12y+c

= 0 at a point Q then coordinates of Q area. (–6, –11) b. (–9, –13) c. (–10. –15) d. (–6, –7)

Solution: Equation of tangent at (1, 7) to the curve y = x2+6 is

6x2

7y+=

+

2x – y+5 = 0 .....(1)This line also touches the circle i.e.Equation of normal of circle passing throughcentre (–8, –6).x+2y + λ = 0–8 –12 + λ = 0

λ = 20020y2x =++∴ .....(2)

Q is intersection point of (1) and (2)x = –6, y = –7Q(–6, –7)Ans: d

132

Exercise: 7 Consider the two curves c1: y2 = 4x, c

2: x2+y2–6x+1 = 0. Then

a. c1 and c

2 touch each other only at one point.

b. c1 and c

2 touch each other only at two points.

c. c1 and c

2 intersect (but do not touch) at exactly two points.

d. c1 and c

2 neither intersect nor touch each other.

Solution: Let eqn of tangent of parabola be

y = m

1mx + is also a tangent to the circle then

22m1m1

m3

2=

+

+

( ) ( )22

22

m18m

1m3 +=+

m4 –2m2+1 = 0m2 = 1 m = ± 1 Two common tangents are possible.

Exercise: 8 If b, c are intercepts of a focal chord of the parabola y2 = 4ax then c is equal to

a. a–b

bb. a–b

ac. b–a

abd. a–b

ab

Solution: We know that SBSA

xSAxSB2a2

+=

acbcabcb

bca −=⇒

+= 2

2

ab+ac = bc

cab

ab =–

Ans: dExercise: 9 The circle x2+y2–2x–6y+2 = 0 intersects the parabola y2 = 8x orthogonally at the

point P. The equation of the tangent to the parabola at P can bea. 2x –y+1 = 0 b. 2x+y –2 = 0c. x+y –4 = 0 d. x –y –4 = 0

Solution: Letm

2mxy += be tangent to y2 = 8x. Since circle intersects the parabola

orthogonally. So this tangent is the normal for the circle. Every normal of the circle passesthrough its centre. So centre (1, 3).

cobab )( −=

133

02m3–mm

2m3 2 =+ +=

(m – 2) (m – 1) = 0m = 1, 2.y = x+2 or y = 2x+1Ans: a

135

PARABOLA - VI

Equation of Normal(i) Point form :

y2 = 4axDifferentiate w.r.t.x

2ydxdy

= 4a

dxdy

= y

a2

Slope of normal = – a2

y1

Equation of normas at (x1,y

1) is

y–y1 = –

a2

y1 (x–x1)

(ii) Parametric form : P(at2,2at)replace x

1 by at2 and y

1 by 2at

y–2at = – a2at2

(x–at2)

y = –tx+at3+2attx+y–at3–2at = 0

(iii) Slope form:Replace t by – my = mx–2am–am3

y = mx+c is normal to parabola y2 = 4ax ifc = –2am–am3 ie., condition of normal.

Equation of NormalParabola Point form Pt.of contact Parametric form Point of contact slope Form Pt.of contact

y2=4ax y–y1=

a2

y– 1 (x–x1) (x

1,y

1) y=–tx+2at+at3 (at2,2at) y=mx–2am–am3 (am2,–2am)

y2= –4ax y–y1=

a2

y1 (x–x1) (x

1,y

1) y=–tx+2at+at3 (–at2,2at) y=mx+2am+am3 (–am2,2am)

x2=4ay x–x1=–

a2x1 (y–y

1) (x

1,y

1) x= –ty+2at+at3 (2at,at2) y=mx+2a+ 2m

a⎟

⎠

⎞

⎜

⎝

⎛

2m

a,

m

a2–

x2=–4ay x–x1=

a2

x1 (y–y1) (x

1,y

1) x= ty+2at+at3 (2at,–at2) y=mx–2a– 2m

a⎟

⎠

⎞

⎜

⎝

⎛

2m

a,–

m

a2

Equation of normal to the parabola (y–k)2= 4(x–h) isy–k = m(x–h) –2am –am3.

P(x ,y )11

Tangent

Normal

136

Properties of Normal1 If the normal at the point P (at

12, 2at

1) meets the parabola at

Q (at2

2, 2at2), then t

2 = –t

1 –

1t

2

Let equation of parabola be y2 = 4ax.Equation of normal at P is

y = – t1x+2at

1+at

13

Point Q lies on the normal, so2at

2 = –at

1t

22+2at

1+at

13

2a(t2–t

1) = –at

1(t

22–t

12)

2 = –t1(t

2+t

1) ( t

1 ≠ t2)

∴ t2 = – t

1 –

1t

2

2 If the normal at the points (at12, 2at

1) and (at

22, 2at

2) meet on the parabola y2 = 4ax, then t

1t2 =2.

Let the equation of normal at (at12,2at

1) and (at

22, 2at

2) be

y = –t1x+2at

1+at

13

and y = – t2x+2at

2+at

23

meet the parabola y2 = 4ax at (at3

2,2at3) then

t3 = –t

1–

1t

2 and t

3 = –t

2 –

2t

2

∴ – t1 –

1t

2= – t

2 –

2t

2

t2–t

1 = 2 ⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛ −=−⇒⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

21

1212

21

211

tt

tttt

tt

⇒ t1t2 = 2

3 No normal other than axis passes through focus.

Let equation of normal be y = mx–2am–am3

passes through (a,0) ie. focus∴ 0 = ma–2am–am3

0 = – am–am3

0= –am((1+m2) ⇒ m = 0 i.e. axis1+m2 = 0 which is not possible.

Ex1 Three normals to y2 = 4x pass through the point (15,12). One of the normals is(A) x+y = 27 (b) x+4y = 63 (c) 3x–y = 33 (d) y+3x = 51Let equation of normal be y = mx–2am–am3. a = 1

∴ y = mx–2m–m3

passes through (15,12)12 = 15m – 2m–m3

y

x

P(at ,2at )1 12

Q(at ,2 )22 at2

137

m3–13m +12 = 0(m–1) (m–3) (m+4) = 0

m = 1, 3, –4.m =1 ⇒ y = x–3m =3 ⇒ y = 3x–33m =–4 ⇒ y + 4x = 72

Ans (c)

Ex2 The minimum distance between the curves x2+y2–12x+31 = 0 and y2 = 4x is

(a) 5 (b) 2 5 (c) 6– 5 (d) None of these.

Centre (6,0) radius = 31–036 + = 5Minimum distance obtained along the common normal.

y2 = 4xDifferentiate w.r.t.x

2y dx

dy= 4

dx

dy = y

2

slope of normal at (x1,y

1) is –

2y1

Also slope of CQ = 6–x

0–y

1

1 = –

2

y1

⇒ y1 = 0 or x

1 = 4

Points are (0,0), (4,4), (4,–4)OC = 6

QC = 2 5

RC = 2 5

Minimum distance ⎪

⎪

⎭

⎪⎪

⎬

⎫

⎪

⎪

⎩

⎪⎪

⎨

⎧

==

==

=

55–52R–RC

55–52R–QC

5–6R–OC

is 5

Ans (a)

138

Important Properties :* If the tangent and normal at any point ‘P’ of the parabola intersect the axis at T and N then ST

= SN=SP where S in the focus.

y

x

P

NS

Q

T

* The portion of a tangent to a parabola cut off between the directrix & the curve subtends a rightangle at the focus.

∠ PSQ = 90°* Any tangent to a parabola and the perpendicular on it from the focus meet on the tangent at the

vertex.∠ PQS = 90°

y

x

P

S

Q

* If the tangents at A and B meet in P then PA and PB subtends equal angles at the focus S.(SP)2 = SA×SBΔ SAP~ Δ SPB∠ PSA = ∠ PSB.

A

B

PS

* The area of the triangle formed by three points on a parabola is twice the area of the triangle formedby the tangents at these points.

139

PRACTICE QUESTIONS

1. The normal at the point (2, 4) of the parabola y2 = 8x meets the parabola at the point

a. (18, –12) b. (12, – 18) c. (12, 18) d. (18, 12)

2. If y+b = m1 (x+a) and y+b = m

2 (x+a) are two tangents to the parabola y2 = 4ax, then

a. m1m

2 = 1 b. m

1m

2 = –1 c. m

1+m

2 = 0 d. m

1– m

2 = 0

3. If normals at the ends of the double ordinate x = 4 of parabola y2 = 4x meet the curve againin P and P′ respectively, then P

P′

=

a. 10 b. 6 c. 12 d. 18

4. Radius of the largest circle which passes through the focus of the parabola y2 = 4x andcontained in it, is

a. 2 b. 4 c. 6 d. 8

5. If the normal at (1, 2) on the parabola y2 = 4x meets the parabola again at the point (t2, 2t),then t is

a. 3 b. 1 c. 2 d. –3

6. If x = my+c is a normal to the parabola x2 = 4ay, then c =

a. 2am+am3 b.

3ma

ma2 +

c. –2am–am3 d. 3ma

–ma2

–

ANSWERS

1. a 2. b 3. c 4. b 5. d 6. c

141

PARABOLA - VII

Conormal points:Let P (h,k) be a point and equation of parabola be y2 = 4ax.Equation of normal is

y = mx–2am–am3

If passes through (h,k) sok =mh –2am–am3

am3+2am–mh+k = 0am3+m(2a–h) +k = 0

Suppose m1, m

2, m

3 are the roots of this equation

∴ m1+m

2+m

3 = 0

m1m

2+m

2m

3+m

3m

1 =

a

h–a2

m1. m

2. m

3 = –

a

k

So maximum three normal say PM, PN, PQ drawn through P. Points M, N, Q are calledco-normal points.

� The algebraic sum of ordinates of the conormal points is zero.Let the coordinates of conormal points be M(am

12,–2am

1), N(am

22,–2am

2) and

Q (am3

2, –2am3). The ordinates of these points

y1+y

2+y

3 = – 2am

1 –2am

2 –2am

3

= – 2a(m1+m

2+m

3)

= 0⇒ y

1+y

2+y

3 = 0

� Centroid of the triangle formed by conormal points lies on the axis of parabola.Let coordinates of conormal points be M(x

2,y

1), N(x

2,y

2) Q(x

3,y

3)

Then centroid is ⎟

⎠

⎞

⎜

⎝

⎛ ++++3

yyy,

3

xxx 321321 = ⎟

⎠

⎞

⎜

⎝

⎛ ++0,

3

xxx 321

Since sum of ordinates is zero. Therefore centroid lies on the axis of parabola.� Normal drawn from a point P(h,k) to the parabola are real and distinct if h>2a.

m12+m

22+m

32>0

⇒ (m1+m

2+m

3)2–2(m

1m

2+m

2m

3+m

1m

3) > 0

⇒ 0–a

)h–a2(2>0

⇒ 2a–h<0⇒ h>2aThis shows that position of point(h,k) should be in shaded region.

� Equation of a circle passing through the conormal pointsLet M(am

12,–2am

1), N(am

22, –2am

2) and Q (am

32,–2am

3)be three points on the parabola y2 =

y

x

M

P(h, k)

NQ

x(a,0)S

y

(2a,0)

142

4ax. These three normals passes through point p(h,k)∴ am3+(2a–h)m+k = 0________________(1)⇒ m

1+m

2+m

3 = 0

m1m

2+m

2m

3+m

1m

3 =

a

h–a2

m1 m

2 m

3 = –

a

k

Let equation of circle be x2+y2+2gx+2fy+c = 0If the point (am2,–2am) lies on the circle then(am2)2+(–2am)2+2g(am2)+2f(–2am)+c = 0a2m4+4a2m2+2gam2–4afm+c = 0a2m4+(4a2m2+2ga)m2–4afm+c = 0_______________ _(2)This equation has four roots say m

1, m

2, m

3, m

4 such

that the circle passes through the points M(am12,–2am

1)

N(am2

2,–2am2), Q(am

32,–2am

3)and S(am

42,–2am

4)

∴ m1+m

2+m

3+m

4= 0 (From equation (2))

0 + m4

= 0 (Since m1+m

2+m

3 = 0)

∴ m4

= 0

⇒ S(0,0)Therefore circle passes through origin.

∴ c= 0Now equation (2) is

a2m4+(4a2+2ga)m2–4afm = 0 ( ÷ am)am3+(4a+2g)m –4f = 0

Now this equation is identical with equation (1)

a

a = g2a4

h–a2

+ = f4–

k

⇒ 2g = – (2a+h), 2f = – 2

k

Equation of circle is

x2+y2 – (2a+h) x – 2

k y = 0

x

M

P(h, k)

QN

O

y

143

PARABOLA - VIIIChord of Contact

Let PA and PB be tangents drawn through the point P (h,k).Equation of tangent at A is

yy1 = 2a (x+x

1)

Equation of tangent at B isyy

2=2a(x+x

2)

Both lines pass through (h,k)∴ky

1 = 2a (h+x

1)_____________(1)

ky2 = 2a (h+x

2)_______________(2)

Hence A(x1,y

1) and B(x

2,y

2) lie on

ky = 2a(x+h) or T = 0ie. equation of chord AB.

Equation of chord whose midpoint (x1,y1) is given :S ≡ y2 –4axS

1 ≡ y12–4ax

1

T ≡ yy1 –2a(x+x

1)

Equation of AB is T = S1

Reflection Property of Parabola Module – 7Let PQ be tangent of the parabola y2 = ax at point P(at2,2at)

Equation of tangent at P is ty = x+at2

Q is the point of intersection of tangent and x–axisSo Q (–at2,0)∴SQ = SA+AQ = a+at2 = a(1+t2)SP = PM = at2+a = a(1+t2)∴SP = SQ∠ SPQ = ∠ SQPAlso, ∠ MPQ = ∠ PQS (alternate angles)

Hence QPSMPQ ∠=∠Also NPSIPN ∠=∠

So PN normal bisects IPS∠

Therefore PI is incident ray then PS is reflected ray. So any ray incident parallel to axis of theparabola after reflection it passes through focus.

Example – 1 If three normals can be drawn to the parabolay2–2y = 4x–9 from the point (a, b), then the range ofthe value of a isa. (2, ∞ ) b. (1, ∞ )c. (– ∞ ,–2) d. (4, ∞ )

Solution: y2–2y = 4x–9

(x ,y )11

(x ,y )22

y

xO

AP(h, k)

B

(x ,y )11

y

xO

P

y

xA

Q

MI

N

S(a, 0)

(3,1)S

x

y

(3,1) (4,1)

144

(y – 1)2 = 4(x – 2)For real three normal’s as h > 2aa > 4Ans: d

Example 2 A circle and a parabola y2 = 4ax intersect at four points. The algebraic sum of theordinates of the four points isa. 0 b. 1 c. –1 d. 4

Solution: Let equation of circle be x2+y2+2gx+2fy+C = 0and equation of parabola is x = at2, y = 2atSolving it, we get a2t4+4a2t2+2agt2+4aft+c = 0

a2t4+2a(2a + g)t2+4aft+c = 0t1+t

2+t

3+t

4 = 0

Or 2at1+2at

2+2at

3+2at

4 = 0

y1+y

2+y

3+y

4 = 0

Ans: a

Example: 3 Maximum number of common normals of y2 = 4ax and x2 = 4by isa. 3 b. 4 c. 6 d. 5

Solution: Normals of y2 = 4ax and x2 = 4by in slope form arey = mx –2am –am2

y = mx + 2b + 2m

b

For common normal 2b + 2m

b = –2am –am2

am4+2am3+2bm2+b = 0This mean there can be atmost 4 common normalsAns: b

Example: 4 The curve described parametrically by x = t2+t+1, y = t2–t+1 representsa. a pair of straight lines b. an ellipsec. a parabola d. a hyperbolaSolution: x+y = 2(t2+1) and x–y = 2t

12

y–x2

yx2

+

=+∴

x2+y2–2xy+4 = 2x+2yx2+y2–2xy–2x–2y+4 = 0

41–1–

1–11–

1–1–1

Δ

��

��

��

= =3–5–2 = 04– ≠

145

h2–ab = 1–1 = 0 ∴ It is a parabolaAns: c

Example: 5 If x+y = k is normal to y2 =12x, then k isa. 3 b. 9 c. –9 d. –3

Solution: Normal to y2 = 12x is y = mx–6m–3m3

m = –1y = –x+6+3x+y = 9

Ans: b

Example: 6 The equation of the common tangent touching the circle (x–3)2+y2 = 9 and the parabolay2 = 4x above the x-axis is

a. 1x3y3 += b. ( )3x–y3 +=

c. 3xy3 += d. ( )1x3–y3 +=

Solution: Equation of tangent to the parabola is y = mx +m1

and

equation of tangent to the circle is y = m(x–3) 2m13 +±

both the equation are identical i.e.2m13m3–

m

1 +±=

( )222

m13m3m

1 +=

+

222 m996m9

m

1+=++

1=3m2 3

1m ±=

Equation of common tangent is 3xy3 += (tangent lying above x–axis)

Ans: c

147

PARABOLA -IXExercise: 1 A tangent PT is drawn at the point P(16, 16) to the parabola y2 = 16x. PT tangent

intersect the x-axis at T. If S be the focus of the parabola, then TPS∠ is equal to

a.2

1tan 1– b.

4

π

c.2

1tan

2

1 1– d.4

3tan 1–

Solution: ST = 4 + AT = 16+4 = 20PS = 4+16 = 20

TPSΔ is isosceles triangle

θtan–1

θtan2

3

4

4–16

0–16θ2tan 2===

02–θtan3θtan2 2 =+( )( ) 02tan1–tan2 =+θθ

2

1θtan = , 2–θtan = (Not possible)

2

1tanθ

1–= (θ is acute angle)

Ans: a

Exercise: 2 If a, b, c are distinct positive real numbers such that the parabolas y2 = 4ax and y2

= 4c (x– b) will have a common normal, then

a. 1c–a

b0 << b. 0

c–a

b <

c. 2c–a

b1 << d. 2

c–a

b >

Solution: Equation of normals arey = mx –2am – am3 ......(1)y = m(x–b) –2cm – cm3 ......(2)Equation 1 and 2 are identical then–2am –am3 = –bm –2cm –cm3 m÷2a+am2 = b+2c+cm2

(a– c)m2 = b+2(c –a)

m2 = 2–c–a

b

2–c–a

bm ±=

S (4,0)θ

θ θ2

148

For m be real 2c–a

b>

Ans: d

Exercise: 3 If AB be a chord of the parabola y2 = 4ax with vertex at A. BC is perpendicular toAB such that it meets the axis at C. The projection of the BC on the axis of parabola isa. 2a b. 4a c. 8a d. 16a

Solution: Let coordinates of B be (x, y)

In x

y

AD

BDθtan,ABDΔ ==

In ( )DC

BDθ–90tan,BCDΔ =

a4x

ax4

x

y.yDC ===∴

Ans: b

Exercise: 4 A circle is descirbed whose centre is the vertex and whose diameter is three-quartersof the latus rectum of the parabola y2 = 4ax. If PQ is the common-chord of the circle and theparabola and L

1L

2 is the latus rectum, then the area of the trapezium PL

1L

2Q is

a.2a

2

22

+b. 4a2 c. 2a22 d. 2a23

Solution: Centre of circle (0, 0)

diameter = a3a4.4

3 =

Eqn of circle 4

a9yx

222 =+ .....(1)

Eqn of parabola y2 = 4ax .....(2)coordinates of P and Q, we get after solving (1) and (2).

4

a9ax4x

22 =+

(x+2a)2 = 2

a5a2–x

2

a52

±=

)possiblenot(2

a9–,

2

ax �=

D

P

B

O

L1

L2q

2a23

Ao

C

B

D90-Oo

x

(x,y)

149

a2,–2

aQ,a2,

2

aP a2y ±=

PQ = 2 a2 , L1L

2 = 4a

Area of trapezium = ( )21LLPQ2

1 + x distance between them.

( )

+=

2

a–axa4a22

2

1

( ) 2a2

22 +=

Ans: a

Exercise: 5 From the point (15, 12) three normals ae drawn to the parabola y2 = 4x, then centroidof triangle formed by three co-normal points is

a. (5, 0) b. (5, 4) c. (9, 0) d.

0,3

26

Solution: Let equation of normal be y = –tx+2t+t3

It passes through (15, 12). So 12 = –15t+2t+t3

t3 –13t –12 = 0(t + 1) ( t+ 3) (t – 4) = 0t = –1, –3, 4

Points are (at2, 2at) i.e. (1, –2), (9, –6), (16, 8)

Centroid is

=

+++

0,3

26

3

86–2–,

3

1691

Ans: d

Exercise: 6 A ray of light travels along a line y = 4 and strikesthe surface of a curve y2 = 4(x+y) then equation of theline along reflected ray travel isa. x+1 = 0 b. y–2 = 0c. x = 0 d. x–2 = 0Solution: y2–4y = 4x(y–2)2 = 4(x+1)Focus (0,2)

Incident ray is parallel to axis of the parabola, so reflected ray passes through focus (0,2) i.e. x = 0

Exercise: 7 Let P be a point of the parabola y2 = 3(2x–3) and M is the foot of perpendicular drawn fromP on the directrix of the parabola, then length of each side of an equilateral triangle SMP, where S isfocus of the parabola isa. 6 b. 8 c. 10 d. 11

(D) x-2=0

P

y

(-1,2)

xO

y'

(0,2)

x'

y=4

150

Solution: Equation of parabola is

y2=6

2

3–x

focus S(3,0)equation of directrix x = 0

P

+ t3,t2

3

2

3 2

Coordinates of M (0,3t)

MS = 2t99 +

MP = 2t2

3

2

3+ , But MS = MP

9+9t2 =22 t

49

t49

49 ++

t2 = 3Length of side = 6

Ans: aPRACTICE QUESTIONS

1. The point (2a, a) lies inside the region bounded by the parabola x2=4y and its latus rectum. Thena. 0 < a ≤ 1 b. 0 < a < 1 c. 0 ≤ a ≤ 1 d. a < 1

2. Two perpendicular tangents to y2=4ax always intersect on the linea. x+a=0 b. x–a=0 c. x+2a=0 d. y+2a=0

3. C1: y2=8x and C

2:x2+y2=2. Then

a. C1 and C

2 have only two common tangents which are mutually perpendicular

b. C1 and C

2 have two common tangents which are parallel to each other.

c. does not have any common tangent.d. C

1 and C

2 have four common tangents.

4. Two common tangents to the circle x2+y2=2a2 and y2=8ax area. y=

±

(x+a) b. x= ± (y+2a) c. y= ± (x+2a) d. x= ± (y+a)5. The number of points with integral coordinates that lie in the interior of the circle x2+y2=16 and the

parabola y2=4x area. 6 b. 8 c. 10 d. 12

6. The vertex of the parabola x2+y2–2xy –4x+4=0 is at

a. ⎟

⎠

⎞

⎜

⎝

⎛

2

1,–

2

1– b. (–1, –1) c. (1, 1) d. ⎟

⎠

⎞

⎜

⎝

⎛+2

1,

2

1

7. The length of the latus rectum of the parabola 2{(x–a)2 + (y–a)2}=(x+y)2 is

a. 2 a b. 2a c. 2 2 a d. 3 2 a

8. The point on y2=4ax nearest to the focus is

M

P

y'

S

y

x' xA

151

a. (0, 0) b. (a, 2a) c. (a, –2a) d. (4

a, a)

9. The angle between the tangents drawn from the origin to the parabola y2=4a(x–a) is

a. 45° b. 60° c. 90° d. tan–1

2

1

10. The circle x2+y2+2λx=0, λεR, touches the parabola y2=4x externally. Thena. λ > 0 b. λ < 0 c. λ > 1 d. none of these

ANSWERS1. b 2. a 3. a 4. c 5. b6. d 7. c 8. a 9. c 10. a

153

PARABOLA - XExample: 1 If the chord of contact of tangent from a point P to the parabola y2 = 4ax touches the parabola

x2 = 4by. The locus of P isa. Parabola b. Hyperbola c. ellipse d. Circle

Solution: Let the point P be (h,k) then equation of chord of contact is ky = 2a(x+h).................(1)Now this chord is tangent of parabola x2 = 4by x2 = 4byEquation of tangent xx

1-2 by

1 = 2by ------------------(2)

From (1) and (2) we get

So, Equation of parabola becomes

4ab = –hkLocus of (h,

,k) is xy = –2ab. i.e. Hyperbola.

Ans: b . xy= 4ab

Example: 2 Let P and Q be points (4, –4) and (9, 6) on the parabola y2 = 4a(x–b). R is a point on the

parabola so that area

PRQΔ

is maximum, then

a. 090PRQ =∠ b. the point R is (4,4)

c. the point R is

1,

41

d. None of these

Solution: (4,–4) lies on y2 = 4a(x–b) 16=4a(4–b)(9,6) lies on y2 = 4a(x–b) 36=4a(9–b)

a = 1

∴

y2 = 4xLet R be (t2,2t) on the Parabola thus

area

PRQΔ

=1t2t

169

14–4

21

2 = ( ) ( ){ }22 t6–t18t–94t2–64

21 ++

= ( )22 t6–t18t4–36t8–2421 ++ = ( )60t10t10–

21 2 ++

0bb9b4

3616 = = R

Q (9.6)

(4 -y)1

k

ba4and y =

1

k

ba4x

1=

2

154

= –5(t2–t–6) = –5 45

3021

–t2

++

Area is maximum when t =21

Coordinates of R

1,

4

1

Ans: c

Example: 3 Minimum area of circle which touches the parabola’s y = x2+1 and y = x2-1 is

a. unit.sq32π9

b. unit.sq4

π

c. unit.sq32

π7d. unit.sq

16

π9

Solution: y = x2+1 and y2 = x+1 are symmetrical about y = xtangent at point. A and B are parallel to the line y = xy = x2+1 y2 = x-1

1x2dxdy ==

21

y =

21

x =45

x =

45

y =

45

,21

A

21

,45

B

AB = 4

2321

–45

45

–21

22

=

+

Area of circle = 2rπ = unit.sq32π9

823

π

2

=

Ans: a

Example: 4 The equation of the common tangents to the parabola y = x2 and y = –(x–2)2 is / area. y = 4(x–1) b. y = 0 c. y = –4(x–1) d. y = –30x–50

Solution: Let y = mx+c is tangent to y = x2

∴ mx+c = x2

x2–mx–c = 0 has equal rootsm2+4c = 0

y = mx–

4

m2

is tangent to y = –(x–2)2 also

By=x

A

4

125)2/1(5 +−−=⇒ tA2

y2 = x-1

y = x2+1

155

mx–4

m2

= –x2+4x–4

x2+(m–4)x+4–4

m2

= 0 has equal roots

∴ (m–4)2 –4

04

m–4

2

=

m2+16–8m–16+m2 = 0m2–4m = 0m = 0, 4Equation of tangent are y = 0 and y = 4x–4

Ans: a,b

Example 5.(3,0) is the point from which three normals are drawn to the parabola y2 = 4x which meet theparabola in the points P, Q and R thenColumn I Column IIi. Area of PQRΔ a. 2

ii. Radius of circum circle of PQRΔ b.25

iii. Centroid of PQRΔ c.

0,2

5

iv. Circum centre of PQRΔ d.

0,3

2

Solution: Equation of normal is y = mx–2m–m3

It passes through (3,0), so 3m–2m–m3 = 0m(1–m2) = 0m = 0, 1, –1

Points are given by (m2, –2m)i.e. P(0,0), Q(1,–2), R(1,2)

area of PQRΔ = units.sq2

121

12–1

100

21 =

25

2.44.5.5

Δ4abc

R ===

Centroid PQRΔ =

0,32

R

Q

P

156

Circum centre

0,2

5

Comprehension based Questions (Exampels 6 to 8)Comprehension 1Consider the circle x2+y2 = 9 and the parabola y2 = 8x. They intersect at P and Q in the first and thefourth quadrants, respectively. Tangents to the circle at P and Q intersect the x-axis at R and tangentsto the parabola at P and Q intersect the x-axis at S.

Example 6.The ratio of the area of the triangles PQS and PQR is

a. 2:1 b. 1:2 c. 1:4 d. 1:8

Solution: Point of intersection of circle & parabolax2+8x–9 = 0(x+9) (x–1) = 0X = 1, –9 (not possible)

Y = 52±

P ( )22,1 , Q ( )22,–1

Tangent to the parabola at P is y22 = 4(x+1)

S(–1,0)

Tangent to the circle at P is x+ 9y22 =R(9,0)

4

1

8

2

RT

ST

xPQxRT21

xPQxST21

PQRΔar

PQSΔar ====

Ans: c

Example 7. The radius of the circum circle of the triangle PRS is

a. 5 b. 33 c. 23 d. 32

Solution: area 21022x10x2

1xSRxPT

2

1ΔPRSΔ ====

33210.4

32.26.10

Δ4

abcR ===

Ans: bExample 8.The radius of the in circle of the triangle PQR is

a. 4 b. 3 c.3

8d. 2

S

Q

R R

P

22±

T

157

Solution: r = 2

28

216

2242626

8.24.21

2QPRQPR

PQxRT21

S

Δ ==++

=++

==

Ans: d

COMPREHENSION 2 (EXAMPLES 9 TO 11)If y = x is tangent to the parabola y = ax2+c

9. If a = 2, then the value of c is

a.2

1b.

4

1c.

8

1d. 1

10 If (1,1) is point of contact then ‘a’ is

a. 1 b.2

1c.

3

1d.

4

1

11 If c = 2 then point of contact is

a. (4,4) b. (2,2) c. (8,8) d.

2

1,

2

1

9. Solution: y = ax2+c

∴ thus 8

1c = for a = 2

Ans: c

10. If (1,1) is point of contact then 2

1a =

Ans: b

11. If c = 2, then point of contact is

Since it lies on the line y = x,

8

12

4

1

2

1 =⇒+= aaa

∴ point of contact is (4,4)

PRACTICE QUESTIONS1. The point P on the parabola y2=4ax for which ⏐PR–PQ⏐ is maximum, where R (–a, 0), Q (0, a) is

a. (4a, –4a) b. (4a, 4a) c. (a, 2a) d. (a, –2a)

1ax2dx

dy == Point of contact of the tangent is since it lies on y = xa

x2

1=⇒

+

a4

1,

a2

1⎟

⎠

⎞

⎜

⎝

⎛ c

+ 2

a4

1,

a2

1⎟

⎠

⎞

⎜

⎝

⎛

158

2. The shortest distance between the parabola y2=4x and the circle x2+y2+6x–12y+20=0 is

a. 4 2 –5 b. 4 2 +5 c. 3 2 +5 d. 3 2 –5

3. If normals are drawn from a point p (h, k) to the parabola y2=4ax, then the sum of the interceptswhich the normals act off from the axis of the parabola isa. 4(h+0) b. 3(h+c) c. 2(h+a) d. (h+a)

4. If a ≠ 0 and the line 2px+3qy + 4r =0 passes through the points of intersection of the parabolasy2=4ax and x2=4ay, thena. r2+(3p+2q)2=0 b. r2+(2p+3q)2=0 c. r2+(3p–2q)2=0 d. r2+(2p–2q)2=0

5. The equation of the tangent at the vertex of the parabola x2+4x+2y=0 isa. x=2 b. x=–2 c. y=–2 d. y=2

6. The common tangent to the parabolas y2=4ax and x2=4ax and x2=32ay isa. x+2y–4a=0 b. x+2y+4a=0 c. x–2y+4a=0 d. x–2y–4a=0

7. The shortest distnae between the parabolas y2=4x and y2=2x–6 is

a. 5 b. 2 c. 3 d. none of these

8. The largest value of a for which the circle x2+y2=a2 falls totally in the interior of the parabola y2=4(x+4)is

a. 4 b. 4 3 c. 3 3 d. 2 3Multiple choice questions with one or more than one correct answer.

9. Let P(x1, y

1) and Q(x

2, y

2), y

1<0, y

2 ,0, be the end points of the latus rectum of the ellipse x2+4y=4.

The equations of parabolas with latus rectum PQ are

a. x2+2 3 y=3+ 3 b. x2–2 3 y=3+ 3

c. x2+2 3 y=3– 3 d. x2–2 3 y=3– 310. The tangent PT and the normal PN to the parabola y2=4ax at a point P on it meet its axis at point T

and N, respectively. The locus of the centroid of the triangle PTN is a parabola whose

a. vertex is ⎟

⎠

⎞

⎜

⎝

⎛ 0,3

a2b. directrix is x=0

c. latus rectum is 3

a2d. focus is (a, 0)

Match the following :11. Consider the parabola y2=12x

Column I Column IIA Equation of tangent can be p. 2x+y–6=0B. Equation of normal can be q. x–2y–12=0C. Equation of chord of contact w.r.t. any point on the directrix r. 2x–y=36D. Equation of chord which subtends right angle at the vertex s. 3x–y+1=0

Assertion and Reasoning

12. Statement 1 : The curve y=–2

x2

+x+1 is symmetric with respect to the line x =1.

Statement 2 : A parabola is symmetric about its axis.(A) Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanations

159

for statement 1.(B) Statement 1 is True, statement 2 is true, statement 2 is not a correct explanation

for statement 1.(C) Statement 1 is true, statement 2 is false.(D) Statement 1 is false, statement 2 is true.

ANSWERS1. c 2. a 3 c 4. b 5 d6. b 7. a 8. d 9. b, c 10. a, d11. A → s, B → r, C → p, D → q 12. A

1 straight line-1 12-07-2015 - cbsecontents s.no. topics page number 1. coordinate geometry - i 1 2...

Documents