circles

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Mathematics / Circles

Circles

As we know very well from pure geometry, a circle is a geometrical figure described by a moving point in theEuclidean plane such that its distance from a fixed point is always constant. The fixed point is called the centre ofthe circle while the fixed distance is called the radius of the circle.

Moving over to a co-ordinate system, let us denote the centre C of the circle by (x0, y0) and the radius by r. Forany point P(x, y) lying on the circle, the length PC must be equal to r. Using the distance formula for PC, wetherefore obtain:

( ) ( )2 2 20 0 : Equation of thecirclex x y y r− + − =

This equation must be satisfied by every point P(x, y) lying on the circle; therefore, this is the equation that uniquelydescribes the given circle. We simply call it the equation of the circle, with centre (x0, y0) and radius r.

For example, consider the circle with its centre at (1, 1) and radius equal to unity:

Fig - 01

(1,1)

y

A circle with centre(1 1) and 1, r =

x

The equation of the this circle is

( ) ( )2 2 21 1 1x y− + − =

2 2 2 2 1 0x y x y⇒ + − − + =

Find the equation of the circle passing through the points (0, 0) , (3, 0) and (1, 2).

Section - 1 EQUATIONS DESCRIBING CIRCLES

Example – 01



Solution: To write the equation of the required circle, we must find its centre and radius.

Recall from pure geometry that a circle can always be drawn through three non-collinear points. Thiscan be done as follows: join the points to form a triangle. Draw the perpendicular bisectors of any ofthe two sides of this triangle. Their point of intersection gives us the centre C. The distance of C fromany of the vertices gives the radius r of the circles:

C is the centre of the circle passing

Fig - 02

C

P

Q R

through P, Q, R.

CP = CQ = CR = r.

We apply this result to the current example:

The equation of the perpendicular bisectorof OB is

32

x =

The equation of the perpendicular bisector

y

xO (0,0) B (3,0)

C

A (1, 2)

Fig - 03

of OA is1 11 22

y

x

− = −−

12 22

y x⇒ − = −

52 02

x y⇒ + − =

The point C is the intersection of the two angle bisectors

3 1,2 2

C ⇒ ≡

We can now easily evaluate the radius r as the length OC:

9 1 104 4 2

r OC= = + =

Finally, the equation of the required circle becomes:2 23 1 10

2 2 4x y − + − =

We will subsequently see another method to solve this type of questions.



Find the equation of the circle which touches the co-ordinate axes and whose centre lies on the line 2 3x y− =

Solution: A circle of radius r touching the co-ordinate axes can be in one of the four following configurations,with four corresponding equations mentioned alongside:

Fig - 04

( , )r r

y

x( � ) + x r 2 ( � ) = y r r2 2

(� , )r r

y

x( ) + x + r 2 ( � ) = y r r2 2

( )r – r

y

x

( � ) + x r 2 ( ) = y + r r2 2

(� , )r r�

y

x

( ) + x + r 2 ( ) = y + r r2 2

Note from these four possible cases that the centre of such a circle either lies on y = x or on y = � x.

In the current example, the centre is also given to lie on x � 2y = 3. Thus, there will be two circles,with the two centres being given by the point of intersection y = x and y = � x with x � 2y =3.

( )1and 2 3 3,3y x x y C= − = ⇒ ≡ −

⇒ Equation of the circle is ( ) ( )2 23 3 9x y+ + − =

( )2and 2 3 1, 1y x x y C= − − = ⇒ ≡ −

⇒ Equation of the circle is ( ) ( )2 21 1 1x y− + + =

Find the equation of the circle with radius 5 and which touches another circle 2 2 2 4 20 0x y x y+ − − − = externallyat the point (5, 5)

Solution: Let us first try to rearrange the equation of the given circle in the standard form from which we�ll beable to deduce its centre and radius:

2 2 2 4 20 0x y x y+ − − − =

( ) ( )2 21 2 25x y⇒ − + − =

Example – 02

Example – 03



Therefore, the centre of this circle is (1, 2) and its radius is 5.

We should now draw a geometrical figure which will certainly make things more clear:

Fig - 05

y

x

( )x , y0 0

(5, 5)

(1, 2)

5

The required circle

Let the centre of the required circle be ( , ). Observe that is the mid-pt of (1,2) and ( , )

x yP

x y

0 0

0 0P

The given circle

As explained in the figure, we can now evaluate (x0, y0), the centre of the required circle:

00

1 5 92

x x+ = ⇒ = 00

2 5 82

y y+ = ⇒ =

Thus, the required equation is

( ) ( )2 29 8 25x y− + − =

EQUATION OF A CIRCLE : GENERAL FORM

Expanding the standard form of the equation of the circle we derived in the last section, we�ll obtain:

2 2 2 2 20 0 0 02 2 0x y x x y y x y r+ − − + + − =

This suggests that the most general form of the equation of a circle can be written in terms of three variables; callthem g, f and c so that

2 2 20 0 0 02 2 ; 2 2 ;g x f y c x y r= − = − = + −

2 2 2 2 20 0 0 0; ;x g y f r x y c g f c⇒ = − = − = + − = + −

Thus, the equation of the circle in terms of g, f and c becomes

2 2 2 2 0x y gx fy c+ + + + = : Equation of a circle; most general form

( ) 2 2Centre , ; Radiusg f g f c= − − = + −

It should be apparent to you how the standard form and the general form of the circle�s equation are interconvertible.Which form to use where is a matter of convenience and will depend on the situation.



As a first example, let us redo Example - 1, which involves finding the equation of the circle passing through thepoints (0, 0), (3, 0) and (1, 2).

Let the equation be 2 2 2 2 0x y gx fy c+ + + + = , where g, f and c are to be determined. This equation must besatisfied by the three points through which the circle passes, and hence we�ll obtain three equations from whichg, f and c can be determined:

Substitute (0, 0) : c = 0

Substitute (3, 0) : 9 + 6g = 0

32

g⇒ = −

Substitute (1, 2) : 1 + 4 � 3 + 4f = 0

12

f⇒ = −

The required equation is hence:

2 2 3 0x y x y+ − − =

which is the same as what we obtained in Example - 1.

Let C be any circle with centre ( )0, 2 . Prove that at the most two rational points can lie on C.

Solution: By a rational point, we mean a point which has both its co-ordinates rational.

Let the equation of C be 2 2 2 2 0x y gx fy c+ + + + =

We can arrive at the result easily be contradiction. Suppose that we have three rational points on thecircle with the co-ordinates (xi, yi) i = 1, 2, 3. These three points must satisfy the equation of thecircle. Thus we obtain a system of linear equations in g and f :

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

2 22 2 1 1 1 11 1 1 12 2 2 22 2 2 2 2 2 2 22 2

2 23 3 3 33 3 3 3

2 22 2 02 2 0 2 22 2 0 2 2

x g y f c x yx y gx fy cx y gx fy c x g y f c x yx y gx fy c x g y f c x y

+ + = − ++ + + + = ⇒ + + + + = ⇒ + + = − + + + + + = ⇒ + + = − +

The coefficients in this system of linear equations are all rational by assumption. Thus, when we solve

this system, we must obtain g, f and c to be all rational. But since the centre is ( )0, 2 , we have

2f = − which gives us a contradiction.

This means that our assumption of taking three rational points on the circle is wrong ⇒ At the mosttwo rational points can lie on this circle.

Example – 04



Suppose we are given two curves C1 and C2 whose equation are as follows:

2 21 1 1 1 1 1 1: 2 2 2 0C a x h xy b y g x f y c+ + + + + =

2 22 2 2 2 2 2 2: 2 2 2 0C a x h xy b y g x f y c+ + + + + =

It is also given that these curves intersect in four concyclic points. Prove that

1 1 2 2

1 2

a b a bh h− −=

Solution: From the discussions in the last chapter, we know that any curve C passing through the point(s) ofintersection of two given curves 1 20 and 0C C= = can be written as

1 2 0 whereC C C≡ + λ = λ ∈ !

We can do the same in the current example to obtain the equation of the curve passing through the four(concyclic) points of intersection as:

( ) ( ) ( ) ( ) ( )2 21 2 1 2 1 2 1 2 1 2 1 22 2 2 0a a x h h xy b b y g g x f f y c c+ λ + +λ + +λ + + λ + + λ + + λ =

From the general form of the equation of the circle, we know that this equation (above) will representthe equation of a circle only if:

Coeff. of x2 = Coeff. of y21 2 1 2a a b b⇒ + λ = + λ

1 1

2 2

b ab a−⇒ λ = −− ... (1)

Coeff. of xy = 0 1 2 0h h⇒ + λ =

1

2

hh

⇒ λ = − ... (2)

From (1) and (2), we have

1 1 2 2

1 2

b a b ah h− −=

Suppose that the equation of a circle is

2 2 2 2 0S x y gx fy c≡ + + + + =

What condition must the co-ordinates of a point P(x1, y1) satisfy so that P may lie (i) inside the circle (ii) outside thecircle?

Example – 05

Example – 06



Solution: Let the centre of S be C and its radius be r.

The point P lies inside S if CP < r and outside S is CP > r.

From the equation of S, we know C to be (�g, � f) and r to be 2 2 .g f c+ − Using these facts, wecan easily evaluate the required conditions:

P lies inside the circle: 2 2CP r<

( ) ( )2 2 2 21 1x g y f g f c+ + + < + −

2 21 1 1 12 2 0x y gx fy c+ + + + < ... (1)

P lies outside the circle: 2 2CP r>

( ) ( )2 2 2 21 1x g y f g f c+ + + > + −

2 21 1 1 12 2 0x y gx fy c+ + + + > ... (2)

We can write (1) and (2) concisely as

( )( )( )

1 1

1 1

1 1

lies inside the circle , 0

lies on the circle , 0

lies outside the circle , 0

P S x y

P S x y

P S x y

⇒ <

⇒ =

⇒ >

Find the equation of the circle circumscribing the triangle formed by the lines 6, 2 4 and 2 5x y x y x y+ = + = + = .

Solution: Let us first consider the general case wherein we�ve been given three lines L1, L2 and L3 and we needto find the circle circumscribing the triangle that these three lines form

Fig - 06

O

C

BA

L3L2

L1

Example – 07



One way to do it would be as follows:

! Find the intersection points A, B and C of the three lines! Use these intersection points to write any two perpendicular bisectors! Find the intersection of these two perpendicular bisectors which gives us the centre O.! Finally, find the radius (which will equal OA, OB and OC)

This procedure will definitely become quite lengthy. We look instead for a a more elegant method.We first try to write the equation of an arbitrary second-degree curve S passing through the intersectionpoints of L1, L2 and L3. Think carefully and you�ll realise that such a curve can be written in terms oftwo arbitrary constants λ and µ as follows:

1 2 2 3 3 1 0S L L L L L L≡ + λ +µ =

That such a curve S will pass through all the three intersection points can be verified by observing thatthe substitution of the co-ordinates of any of the three points in the equation above will make bothsides identically 0.One we have such a curve, we can impose the necessary constraints to make it a circle.Coming back to the current example, the equation of an arbitrary curve passing through the intersectionpoints of the three lines can be written as:

( )( ) ( )( ) ( )( )6 2 4 2 4 2 5 2 5 6 0S x y x y x y x y x y x y≡ + − + − + λ + − + − +µ + − + − =

To make S the equation of a circle, we simple impose the following constraints:2 2 2 1 2 2= ⇒ + λ + µ = + λ + µ2Coeff. of Coeff. ofx y

1⇒ µ = .. (1)

3 5 3 0⇒ + λ + µ =Coeff. of = 0xy

65⇒ λ = − ... (2)

We substitute λ and µ back into S to obtain the required equation as:2 2 17 19 50 0S x y x y≡ + − − + =

As another example of following such an approach, suppose that we are given four straight lines andare told that they intersect at four concyclic points, as shown below:

D

L2

Fig - 07

L4

L1

L3

C

B

A

A, B, C, D are four concyclic points



What approach will you follow if you�re told to find the equation of the circle circumscribing thisquadrilateral? Obviously, one can always proceed by explicitly determining the centre and the radiusof the said circle, but as in the previous question, a much more elegant method exists.

Convince yourself that any second degree curve S passing through A, B, C, D can we written as

1 3 2 4 0S L L L L≡ + λ =

Observe carefully that the substitution of the co-ordinates of any of the four points A, B, C, D willmake both sides identically 0, implying that these four points lie on S. We now simply impose the

necessary constraint ( )onλ to make S represent a circle, thus obtaining S !

Consider a circle of radius r centred at the origin:

2 2 2x y r+ =

A line y = mx + c either just touches this circle or intersects it in two distinct points. What condition must m and csatisfy?

Solution: What we need to do algebraically is solve the simultaneous system of equations

2 2 2x y r+ = ... (1)

y mx c= + ... (2)

and find the condition on m and c for this system to have two distinct roots.

We substitute the value of y from (2) in (1):

( )22 2x mx c r+ + =

( )2 2 2 21 2 0m x mc x c r⇒ + + + − = ... (3)

Since the line intersects the circle (or touches it) the discriminant of (3) cannot be non-negative sinceat least one real value of x must exist. Thus:

( )( )2 2 2 2 24 4 1m c m c r≥ + −

2 2 2 2 0c r m r⇒ − − ≤

( )2 2 21c r m⇒ ≤ +

This is the condition that m and c must satisfy.Incidentally, we also obtained the condition for tangency:

( )2 2 21c r m= +

21c r m⇒ = ± +

Example – 08



Thus, 21y mx r m= ± + will always be tangents to the circle x2 + y2 = r2, whatever be the value ofm. A particular case that you should observe here is that when ,m →∞ the equation becomes

2 2lim

1 1m

y m x rm m→∞

= ±

+ +

x r⇒ = ±

which means that the two tangents are vertical and touch the circle at the two end-points of thehorizontal diameter. This is intuitively obvious.

A point P moves in the Euclidean plane in such a way that ,PA PB= λ where A and B are fixed points and 0λ > .Find the locus of P.

Solution: The easiest case is when 1;λ = then PA = PB and P will hence lie on the perpendicular bisector of AB.We consider the case when 1.λ ≠ Let A and B be assigned the co-ordinates (a, 0) and (� a, 0) (forconvenience). This can always be done by an appropriate choice of the co-ordinate axes.

Now, let P have the co-ordinates (x, y). We have,

2 2 2PA PB= λ

( ) ( ){ }2 22 2 2x a y x a y⇒ − + = λ + +

( ) ( ) ( ) ( )2 2 2 2 2 2 21 1 2 1 1 0x y ax x a⇒ − λ + − λ − + λ + −λ =

( )( )

22 2 2

2

12 0

1x y a x a

+ λ⇒ + − + =

−λ ... (1)

This is obviously the equation of a circle centred at ( )( )

2

2

1,0

1

a + λ − λ

. Note that this circle does not

pass through either A or B.

Let us consider an example of this. Let AB be 2 units, so that we can assign (1, 0) and (�1, 0) as theco-ordinates A and B. Let P move in such a way that PA = 2PB, i.e, 2.λ = From (1), the locus of Pis the circle:

( )2 2 2 1 41 0

1 4x y x

++ − + =

−

2 2 10 1 03

x y x⇒ + + + =

Example – 09



The centre of this circle is 5 ,03

− and its radius is ( )

225 40 1

3 3 − + − =

Fig - 08

y

A

P

(1, 0)B

(�1, 0)5 ,03

−

For any point taken on the circumference of this circle, we will have 2

P

PA = PB

The circle that we obtained

A fixed line L1 intersects the co-ordinate axes at P(a, 0) and Q (0, b). A variable line L2 , perpendicular to L1,intersects the axes at R and S. Show that the locus of the points of intersection of PS and QR is a circle.

Solution: The equation of L1, using intercept form, can be written as

1x ya b+ =

bx ay ab⇒ + =

Since L2 is perpendicular to L1, its equation can be written as

2 0L ax by≡ − + λ =

where λ is a real parameter.

Using the equation of L2, we can determine R and S to be ,0aλ −

and 0,bλ

respectively.

Fig - 09

y

xP a, ( 0)

Q , b(0 )

Raλ

− Sbλ0,

L2

Points and are fixed whereas and will vary as varies

P QR S

λ

,0

Example – 10



We now write the equations to PS and QR using the two-point form:

: yPS x aby ax a ab

−λ= ⇒ λ + = λ− .... (1)

: y b abQR abx y bx− = ⇒ − + λ = λ

λ .... (2)

The relation that the intersection point of PS and QR, will satisfy can be evaluated by eliminating λfrom (1) and (2). We thus obtain

aby abxa x y b

λ = =− −

2 2 2 2aby ab y a bx abx⇒ − = −2 2 0x y ax by⇒ + − − =

This represents a circle centred at ,2 2a b

and passing through the origin.

Let 1, , 1, 2, 3, 4ii

m im

=

be four distinct points lying on a circle. Prove that 1 2 3 4 1m m m m =

Solution: We first assume an equation for this circle C, in its general form:

2 2: 2 2 0C x y gx fy c+ + + + =

Since 1,i

i

mm

satisfies the equation of C for i = 1, 2, 3, 4 we have

22

1 22 0 1, 2,3,4i ii i

fm gm c im m

+ + + + = =

4 3 22 2 1 0 1, 2, 3, 4i i i im gm cm fm i⇒ + + + + = =

This last equation tells us that 'im s are the roots of the following equation in m:

4 3 22 2 1 0m gm cm fm+ + + + = : Roots of this equation arem, i = 1, 2, 3, 4

The product of the roots, which is 1 2 3 4m m m m , can easily be seen to be 1 from this equation.

Example – 11



Find the equation of the circle C which has two fixed points ( ) ( )1 1 2 2, andA x y B x y as the end-points of itsdiameter.

Solution: To evaluate the required equation, we can use a well known result from plane geometry: the angle ina semicircle is a right angle.

Fig - 10

B x y( , )2 2

For any point ( , ) onthe circumference of thecircle, the angle isa right angle

P x y

APB

P x y( , )

A x y( , )1 1

Thus, we can use this fact:

(Slope of AP) × (Slope of PB) = � 1

1 2

1 2

1y y y yx x x x− −⇒ × = −− −

( )( ) ( )( )1 2 1 2 0x x x x y y y y⇒ − − + − − = ... (1)

This is the required equation that will represent C. Note that we could equivalently have used thePythagoras theorem in APB∆ to evaluate the equation of C:

2 2 2AP PB AB+ =

( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 21 1 2 2 1 2 1 2x x y y x x y y x x y y⇒ − + − + − + − = − + −

( ) ( )2 21 2 1 2 1 2 1 22 2 2 2 2 2x y x x x y y y x x y y⇒ + − + − + = − −

( )( ) ( )( )1 2 1 2 0x x x x y y y y⇒ − − + − − =

which is the same as what we obtained in (1).

Given the circle 2 2: 2 2 0C x y gx fy c+ + + + = , find the intercepts that it makes on the x-axis and y-axis.

Solution: We did a similar case in Example - 8 by solving simultaneously the equation of the circle and the lineon which the intercept is required. Here, we�ll proceed analogously.

Example – 12

Example – 13



x – intercept

Fig - 11

To find the -intercept,we put = 0 in the equation for : + 2 + = 0 ... (1)

This gives two values for (real and distinct, equalor imaginary) and as shown alongside. We need to find | � |

xy

Cx gx c

x

x xx x

2

1 2

1 2

x1 x2 x-axis

C = 0

x-intercept

From (1), we have 1 2 1 22 and .x x g x x c+ = − =

Thus,

( )21 2 1 2 1 24x x x x x x− = + −

22 g c= −

y - intercept

Fig - 12

We put = 0 to get + 2 + = 0 ... (2)

If this has roots , , the lengthof the intercept is

xy fy c

y y

2

1 2

y1

y - axis

y2

y - intercept( )2

1 2 1 2 1 24y y y y y y− = + −

( )( )22 using 2f c= -

Thus, the intercepts are of length 2 22 and 2g c f c− − respectively. Obviously, if 2g c< the

circle and the x�axis do not touch/intersect; if 2f c< the circle and the y�axis do not touch orintersect.



What values can the variable a take so that the point (a � 1, a + 1) lies inside the circle 2 2 12 12 62 0x y x y+ − + − =

but outside the circle 2 2 8x y+ = .

Solution: We will use the general result that we derived in Example - 6 earlier (refer). Specifically, if S = 0 is theequation of circle and P (x1 y1) be any point, then

( )1 1, 0 lies insideS x y P S< ⇒

( )1 1, 0 lies outsideS x y P S> ⇒

Using these relations for the current case, we obtain

( ) ( ) ( ) ( )2 21 1 12 1 12 1 62 0a a a a− + + − − + + − < ... (1)

22 36 0a⇒ − <

3 2 3 2a⇒ − < < ... (i)

and ( ) ( )2 21 1 8 0a a− + + − > ... (2)

22 6 0a⇒ − >

3 or 3a a⇒ > < − ... (ii)

The intersection of (i) and (ii) gives us the required values of a as

( ) ( )3 2, 3 3, 3 2a ∈ − − ∪

Find the locus of the foot of the perpendicular drawn from the origin upon any chord of a circle2 2 2 2 0S x y gx fy c≡ + + + + = which subtends a right angle at the origin.

Solution: The situation is depicted graphically in the figure below to make things clearer:

Fig - 13

O AB

S

X l m

is the origin. is a particular chord of the circle which subtends a right angle at the origin. We need to find the locusof the foot of the perpendicular, i.e. ( , ).

BO

y

x

X l m( , )

A

Example – 14

Example – 15



Observe that the equation of the chord AB can be written as

( )y m l AB OXx l m− −= ⊥−

∵

2 2lx my l m⇒ + = +

Now, if we homogenize the equation of S using the equation of the chord AB, what we�ll get is theequation of the pair of straight lines OA and OB (as discussed in the last chapter on straight lines). Thisis what we proceed to do:

22 2

2 2 2 2 2

l2 2 0x my lx my lx myx y gx fy cl m l m l m2

+ + + + + + + = + + + ... (1)

This is the joint equation of OA and OB, since OA and OB need to be at right angles, we impose theappropriate constraint for perpendicularity on (1):

2 2Coeff. of coeff .of 0x y+ =

( ) ( )2 2

2 22 2 2 22 2 2 2

2 21 1 0gl cl fm cml m l ml m l m

+ + + + + = + ++ +

( )( )

2 2

22 2 2 2 2 2

2 22 0c l mgl fm

l m l m l m

+⇒ + + + =

+ + +

2 2 02cl m gl fm⇒ + + + + =

This is the equation of a circle. To be more conventional, we should use (x, y) instead of the variablesl and m.

Thus, the required locus is

2 2 02cx y gx fy+ + + + =



Q. 1 Find the greatest and least distances of the point P (10, 7) from the circle 2 2 4 2 20 0x y x y+ − − − =

Q. 2 Through the origin O, a straight line is drawn to cut the line y mx c= + at P. If Q is a point on this line

such that 2OP OQ⋅ = λ , show that the locus of Q is a circle passing through the origin.

Q. 3 What is the area of an equilateral triangle inscribed in the circle 2 2 2 2 0?x y gx fy c+ + + + =

Q. 4 Determine the equation of the circle passing through the points (1, 2) and (3, 4) and touching the line3 3 0.x y+ − =

Q. 5 A circle whose centre is the point of intersection of the lines 2 3 4 0x y− + = and 3 4 5 0x y+ − =passes through the origin. Find its equation.

Q. 6 Find the equation of the circle which touches the x-axis and two of whose diameters lie along2 5 0 and 3 2 8 0x y x y− − = − − =

Q. 7 Find the equation of the circumcircle of an equilateral triangle two of whose vertices are (�1, 0) and(1, 0) and the third vertex lies above the x�axis.

Q. 8 Find the equation of the circle passing through (1,0) and (0,1) and having the smallest possible radius.

Q. 9 The equations of the sides of a quadrilateral are given by ( )0 1, 2,3, 4 .r r r rL a x b y c r= + + = =If the quadrilateral is concyclic, show that

1 3 1 3 1 3 3 1

2 4 2 4 2 4 4 2

a a b b a b a ba a b b a b a b

− +=− +

Q. 10 Find the point on the straight line 2 11y x= + which is nearest to the circle

( )2 216 32 8 50 0x y x y+ + − − =

TRY YOURSELF - I



We first consider the case which we�ve already considered in Example - 8 earlier, namely, when will the liney mx c= + be a tangent to the circle 2 2 2 ?x y a+ =

Earlier, we solved the simultaneous system of two equations and put the discriminant of the resulting quadraticequal to 0 to obtain the condition for tangency. Here, we follow an alternative approach.

We use the simple geometric fact that if a line L is a tangent to a circle S, then the perpendicular distance of thecentre of S from L must equal the radius of S:

Fig - 14

L = 0C

If is tangent to , then must

equal , the radius of .

LS CP

rS

S = 0

P

In the current case, the centre of 2 2 2x y a+ = is (0, 0) and its radius is a. The perpendicular distance of (0, 0)from y mx c= + must equal a, i.e.

21

ca

m=

+

⇒ 2 2 2(1 )c a m= +

This is the same condition that we derived earlier. From this relation, we can infer that any line of the form21y mx a m= ± + will always be a tangent to the circle 2 2 2 ,x y a+ = whatever the value of m may be.

Consider now the problem of finding the equation of the tangent to the circle 2 2 2x y a+ = at a point 1 1( , )P x ylying on the circle.

Fig - 15

We want to find the equation of the tangent at ( , ) lying on the circle

+ =

T P x y

x y a

1 1

2 2 2

P x y( , )1 1

O ,(0 0)T

x

x

Section - 2 TANGENTS AND CHORDS



The tangent T will obviously be perpendicular to OP. Thus, the equation of T is

1 1

1 1

y y xx x y− −=−

2 21 1 1 1xx yy x y⇒ + = + ...(1)

The other piece of information that we have is that 1 1( , )x y lies on the circle and therefore must satisfy its equation.

Thus, 2 2 21 1 .x y a+ = Using this in (1), we obtain the equation of T as

21 1:T xx yy a+ =

If the point 1 1( , )x y has been specified in polar form, i.e. in the form ( cos , sin ),a aθ θ the equation of T becomes

: cos sinT x y aθ θ+ =

We now go to the general case of finding the equation of the tangent to an arbitrary circle2 2 2 2 0S x y gx fy c≡ + + + + = at a point 1 1( )P x y lying on this circle:

Fig - 16

We want to find the equation of the tangent at ( , ) lying on the circle .

T P x yS

1 1

P x y( , )1 1

C g,- f (- )T

x

y

Here again, as in the earlier case, we have two pieces of information which we can put to use :

P lies on S : 2 21 1 1 12 2 0x y gx fy c+ + + + = ...(1)

T is perpendicular to CP : 1 1

1 1

1y y y fx x x g− +× = −− + { }( , ) is any point on x y T

2 21 1 1 1 1 1xx yy gx fy x y gx fy⇒ + + + = + + +

We now add 1 1( )gx fy c+ + on both sides above :



2 21 1 1 1 1 1 1 1( ) ( ) 2 2xx yy g x x f y y c x y gx fy c⇒ + + + + + + = + + + +

Now we use (1) for the RHS above to finally obtain the equation of T as

1 1 1 1: ( ) ( ) 0T xx yy g x x f y y c+ + + + + + =

The expression on the left hand side of the equation above is conventionally denoted as 1 1( , )T x y(Note that T is a function of x and y too). Thus, the equation of the tangent can be written concisely as

1 1 1 1( , ) 0 : Equation of tangent at ( , ) to T x y x y S=

Using an analogous approach, we can write the equation of the normal to the circle2 2 2 2 0x y gx fy c+ + + + = at the point ( )1 1,P x y . You are urged to do this yourself. Note that

every normal of a circle will pass through the circle�s centre.

Find the equation of the tangent to

(a) 2 2 1x y+ = at 1 2,3 3

(b) 2 2 2 4 4 0x y x y+ − − + = at 1 3, 22 2

+ Solution (a) The equation of the tangent to 2 2 2x y a+ = at 1 1( , )x y is, as obtained in the preceeding discussion,

1 1( , ) 0T x y =2

1 1xx yy a⇒ + =

Here, 1 11 2( , ) ,3 3

x y

= and 1a = . Thus, the required equation is 2 3x y+ =

(b) The equation of the tangent to 2 2 2 2 0x y gx fy c+ + + + = at 1 1( , )x y is

1 1( , ) 0T x y =

1 1 1 1( ) ( ) 0xx yy g x x f y y c⇒ + + + + + + =

Here, 1 1( , )x y is 1 3, 22 2

+ and thus, the required equation is

3 1 32 1 2 2 4 02 2 2 2x y x y

+ + − + − + + + =

3 13 02 2 2x y− ⇒ + − + =

3 (2 3 1) 0x y⇒ − + − + =

Example – 16



What is the length of the tangent to 2 2 2 2 0S x y gx fy c≡ + + + + = drawn from an external point 1 1( , )P x y ?

Solution :

Fig - 17

O g f SP x y

PAPB P S

PA PB

(- , - ) is the centre of . ( , ) is an external point.

Note that two tangents, and can drawn from to .

Also, =

1 1

A

O g,-f (- )

B

P x ,y( )1 1

The length PA (or PB) can be evaluated by a simple application of the Pythagoras theorem.

In ,PAO∆ observe that

2 2 2PA PO AO= −

{ } { } 22 2 2 2

1 1( ) ( )x g y f g f c= + + + − + − { } is the radiusAO∵

2 21 1 1 12 2x y gx fy c= + + + + ...(1)

The equation of the circle being represented by 0,S = we can denote the RHS obtained in (1) by

1 1( , ).S x y Thus, the length of the tangent can be written concisely as

1 1( , )PA S x y=

For example, the length of the tangent from (4, 4) to 2 2 2 4 4 0x y x y+ − − + = will be

2 24 4 2 4 4 4 4l = + − × − × +

12=

2 3=

In the next example, we discuss how to write the equation to the pair of tangents PA and PB.

From an external point 1 1( , ),P x y (two) tangents are drawn to the circle 2 2 2 2 0S x y gx fy c≡ + + + + = . Thesetangents touch the circle at A and B. Find the joint equation of PA and PB.

Example – 17

Example – 18



Solution : Consider any point ( , )h k lying on the tangents drawn from P to S.

Fig - 18

Assume ( ) to lie on one of the two tangents or

h, k

PA PB

A

(- )g,-f

B

P x ,y( )1 1

( )h k 1

Since we know two points on the line PA, we can use the two-point form to write its equation :

1

1

y k y kx h x h− −=− −

1 1 1 1( ) ( ) ( ) ( )x y k y x h h y k k x h⇒ − − − = − − −

1 1 1 1( ) ( ) ( ) 0x y k y x h kx hy⇒ − − − + − =

Since PA is a tangent to S, its distance from the centre of the circle, ( , ),g f− − must equal the radius.

This gives

22 21 1 1 1

2 21 1

( ( ) ( ) )( ) ( )

g y k f x h kx hy g f cx h y k

− − + − + − = + −− + − ...(1)

To write the equation of the pair of lines in conventional form, we use ( , )x y instead of ( , )h k in (1),above. Subsequent (lengthy !) rearrangements give:

{ } ( )( )2 2 2 2 21 1 1 1 1 1 1 1( ) ( ) 2 2 2 2xx yy g x x f y y c x y gx fy c x y gx fy c+ + + + + + = + + + + + + + +

The left hand side can be written concisely as 21 1( ( , ))T x y as described earlier whereas the right hand

side can be written concisely as 1 1( , ) ( , ).S x y S x y Thus, the equation to the pair of tangents can bewritten concisely as

21 1 1 1( , ) ( , ) ( , )T x y S x y S x y=

This relation be written in an even shorter form as simply 21T SS= .



Find the equation to the pair of tangents drawn from the origin to the circle 2 2 4 4 7 0x y x y+ − − + = .

Solution: We use the relation obtained in the last example, 21,T SS= to write the desired equation. Here,

1 1( , )x y is (0, 0) while 2, 2g f= − = − and 7c = . Thus the joint equation is

2 (0,0) ( , ) (0,0)T S x y S=

2 2 2( 2 2 7) ( 4 4 7)(7)x y x y x y⇒ − − + = + − − +

2 2 2 24 4 49 8 28 28 7 7 28 28 49x y xy x y x y x y⇒ + + + − − = + − − +

2 23 8 3 0x xy y⇒ − + =

As expected, since the tangents have been drawn from the origin, the obtained equation is a homogenousone.

Refer to Fig - 18. Suppose that A and B, the points of contact of the two tangents, are joined. What will be theequation to this chord of contact ?

Solution: Assume the co-ordinates of A and B to be ( , )A Ax y and ( , )B Bx y respectively. The equations of thetangents at A and B are, respectively,

}( ) ( ) 0( ) ( ) 0

A A A A

B B B B

xx yy g x x f y y cxx yy g x x f y y c

+ + + + + + =+ + + + + + = ...(I)

Since the two tangents intersect at P, the co-ordinates of P must satisfy the system of equations I.Thus,

}1 1 1 1

1 1 1 1

( ) ( ) 0( ) ( ) 0

A A A A

B B B B

x x y y g x x f y y cx x y y g x x f y y c

+ + + + + + =+ + + + + + = ...(II)

Now, if you observe the system II carefully, you will realise that we can think of it this way : ( , )A Ax y

and ( , )B Bx y are two points which satisfy the linear equation

1 1 1 1( ) ( ) 0xx yy g x x f y y c+ + + + + + = ...(1)

This is because substitution of ( , )A Ax y or ( , )B Bx y into (1) results in the system of equations II.Thus (1) must be the equation of the chord of contact we are looking for, since both A and Bsatisfy (1).

Example – 19

Example – 20



We can write this obtained equation concisely as

1 1( , ) 0T x y = : Equation of the chord of contact

As an example, the chord of contact for the two tangents drawn from the origin to the circle2 2 4 4 7 0x y x y+ − − + = will be

1 1 1 1( ) ( ) 0xx yy g x x f y y c+ + + + + + = where 1 1( , ) (0,0)x y =

2 2 7 0x y⇒ − − + =

72

x y⇒ + =

From an external point P, a line is drawn intersecting a circle S in two distinct points A and B. A tangent is alsodrawn from P touching the circle S at T. Prove that PA PB⋅ is always constant, and equal to 2.PT

Solution: Let the equation of the circle be 2 2 2 2 0S x y gx fy c≡ + + + + = and the point P be 1 1( , )x y :

Fig - 19

We need to prove that =PA PB PT�

2

P x ,y( )1 1 BA

S

T

Notice that the line PAB passes through the fixed point P. What is variable about it is its slope, whichwe assume to be tanθ . Thus, using the polar form for the lines, we obtain the equation of PAB as

1 1

cos sinx x y y r

θ θ− −= = ...(1)

In particular, the co-ordinates of A and B can be obtained in terms of θ using the value of r as PA andPB respectively in (1).

Using (1), we can write any point on the line PAB as 1 1( cos , sin ).x r y rθ θ+ + If this point lies on thecircle, it must satisfy the circle�s equation :

2 21 1 1 1( cos ) ( sin ) 2 ( cos ) 2 ( sin ) 0x r y r g x r f y r cθ θ θ θ+ + + + + + + + =

Example – 21



2 2 21 1 1 1(2 cos 2 sin ) 2 2 0r g f r x y gx fy c⇒ + + + + + + + =θ θ ...(2)

The equation (2) in r will have two roots 1r and 2r corresponding to PA and PB since A and B lie onthe circle.

Thus,

2 21 2 1 1 1 12 2PA PB r r x y gx fy c⋅ = ⋅ = + + + +

1 1( , )S x y=

2PT= (using the result obtained in Example - 17)

You are urged to prove this result using �pure� geometry.

From an external fixed point ( , ),P h k tangents are drawn to the circle 2 2 2x y a+ = . Find the area of the triangleformed by these tangents and their chord of contact.

Solution:

Fig - 20

AB

P x y a

PAB

AB PC

is the chord of contactfor the tangents drawn from

to + = Observe that the area of

can be written

as �

2 2 2

∆

� P h,k( )

B

A

C(0 0),

O12

Observe that OAC∆ is similar to OAP∆ . Thus, the ratio of the corresponding sides is equal.

OC AC OAOA AP OP

= =

OA is simply the radius a, while OP is 2 2 .h k+ Thus,

2 2

2 2

OA aOCOP h k

= =+

Example – 22



Now, AC can be evaluated using Pythagoras theorem in :OAC∆

2 2AC OA OC= −

4

22 2

aah k

= −+

2 2 2

2 2

h k aah k+ −=+

The area of PAB∆ is

12

AB PC∆ = × ×

1 (2 ) ( )2

AC OP OC= × × −

2 2 2 3/ 2

2 2

( )a h k ah k+ −=+

____________________________________________________________________________________

Observe that this question was solved mostly through �pure� geometrical considerations. A pure co-ordinateapproach for the area of the triangle can be followed but you can expect it to be much longer. Thus, in this subject,you have to put your intuitive skills to the best use possible to determine the shortest approach.

Let us consider another example here itself which will show why a pure geometrical approach is better sometimesthan using co-ordinates.

Consider two circles with the following equations :

2 21 2 2 0S x y gx fy c≡ + + + + =

2 2 2 2 2 22 2 2 sin ( )cos 0S x y gx fy c g fα α≡ + + + + + + =

These two circles are obviously concentric, centred at ( , ).g f− − The radius of 1S is 2 21r g f c= + − while

that of 2S is 2 22 1sin sinr g f c rα α= + − = so that 1 2.r r>

The problem is as follows : from any point on 1,S two tangents are drawn to 2.S What is the angle between thesetwo tangents.



Obviously, this angle should be the same for any point chosen on 1S since the circles are perfectlysymmetrical figures. Let us draw a diagram corresponding to this situation :

Fig - 21

Note that = = = sin Thus, =

PO rOT r r

OPT

1

2 1 α

⇒

∠∠

α

αTPS = OPT2 = 2

O

T

P

S

As explained in the diagram, the angle between the two tangents can be evaluated by simple geometricconsiderations to be 2 .α Although this example is more or less trivial, the contrast between puregeometric and co-ordinate approach will become more apparent in some subsequent examples.

A point P moves in such a way so that the tangents drawn from it to the circle 2 2 2x y a+ = are perpendicular.Find the locus of P.Solution: To contrast between the various alternatives available to us, we will use all of them here:

A PURE-GEOMETRIC APPROACH:

Fig - 22

T

P

S

O(0,0)

a

From the figure, it is apparent that OTPS is a square since all the angles are right angles andOS = OT = a .Thus, 2OP a= , i.e., the distance of P from O is always 2 ,a i.e., P lies on a circle of radius.

2 a . Thus, P satisfies the equation2 2 22x y a+ =

This circle is called the Director circle of the given circle.

A CO-ORDINATE APPROACH - ILet P be the point ( , ).h k The equation of the pair of tangents drawn from P to the circle is

21T SS=

2 2 2 2 2 2 2 2( ) ( )( )hx ky a x y a h k a⇒ + − = + − + −

Example – 23 DIRECTOR CIRCLE



This combined equation will represent a pair of perpendicular straight lines if

2 2Coeff. of Coeff. of 0x y+ =

2 2 2 2( ) ( ) 0k a h a⇒ − + − =

2 2 22h k a⇒ + =

Using ( , )x y instead of ( , ),h k we obtain the locus of P in conventional form :

2 2 22x y a+ =

A CO-ORDINATE APPROACH - II

The equation of any tangent to the circle 2 2 2x y a+ = can be written as 21 .y mx a m= + +

If this line passes through ( , ),P h k the co-ordinates of P must satisfy this equation:

21k mh a m= + +

2 2 2( ) (1 )k mh a m⇒ − = +

2 2 2 2 2( ) 2 ( ) 0h a m mhk k a⇒ − − + − =

This is a quadratic in m and will yield two values, say 1m and 2 ,m which physically corresponds tothe fact that two tangents can be drawn from ( , )P h k to the circle.

The two tangents are at a right angle if 1 2 1m m = −

2 2

2 2 1k ah a−⇒ = −−

2 2 22h k a⇒ + =

Using ( , )x y instead of ( , )k h k we obtain the locus of P:

2 2 22x y a+ =

If this question were to be encountered in an exam, the pure-geometric approach would certainly turnout to be the fastest !

1C and 2C are two concentric circles, the radius of 2C being twice that of 1C . From a point P on 2,C tangents

PA and PB are drawn to 1.C Prove that the centroid of PAB∆ is on 1.C

Example – 24



Solution: Let us again attempt this question using both a pure-geometric and a co-ordinate approach.PURE-GEOMETRIC APPROACH:

Recall the following straightforward theorem pertaining to right-angle triangles.

D

CB

A

Fig - 23

∆ ABC is right-angled at B and BD is the median drawn from B to the opposite side AC. Then ourtheorem tells us that

BD = AD = CDThis can be proved using simple geometryWe will put this theorem to use in the current example.

P

A

B

D

O

C1

Fig - 24

C2

E

Since the radius r2 of C2 is twice that of C1 (r1) we havePO = 2 OD

⇒ OD = PD. ... (1)Thus,D is the mid-point of PO. This means that in ∆ OAP, AD is the median to OP. By the theoremmentioned above, we have

AD = OD = PD = r1

Thus, in quadrilateral ,ADBO we have 1.AD BD OA OB r= = = = In other words, ADBO is a

parallelogram so E is the midpoint of .OD Thus, 11/ 22

ED r OD= = ...(2)

Also, since ,AE EB= PE is the median of .PAB∆ Thus the centroid lies on .PEFrom (1) and (2), we finally obtain 2 .PD ED= Thus, D divides the median PE in the ratio2 : 1 implying D is the centroid which lies on 1.C



The descriptive nature of the pure-geometric solution just provided might make it appear to be verylong but actually only a few simple elementary geometry facts have been used.

CO-ORDINATE APPROACH

There�s no loss of generality in assuming that the two circles are centred at the origin. Thus, we canwrite their equations as

2 2 21

2 2 22

:: 4

C x y rC x y r

+ =+ =

Assume the point P to have the co-ordinates ( , ).h k The equation of AB (the chord of contact) canthen be written as

Equation of 2:AB hx ky a+ =

We can evaluate the co-ordinates of 1 1( , )A x y and 2 2( , )B x y by simultaneously solving the equationsfor 1C and AB.

Thus, 1x and 2x will be the roots of22

2 2r hxx rk

−+ =

2 2 2 2 2 2 2( ) 2 ( ) 0h k x r hx r r k⇒ + − + − =2

1 2 2 2

2r hx xh k

⇒ + =+

and 2 2 2

1 2 2 2

( )r r kx xh k

−=+

... (3)

1y and 2y will be the roots of22

2 2r ky y rh

− + =

2 2 2 2 2 2 2( ) 2 ( ) 0h k y r ky r r h⇒ + − + − =2

1 2 2 2

2r ky yh k

⇒ + =+

and 2 2 2

1 2 2 2

( )r r hy yh k

−=+

... (4)

If we let ( , )t s be the co-ordinates of the centroid G of ,PAB∆ we have

( ) ( )

21 2

2 2

21 2

2 2

23 13

Using 3 and 423 1

3

x x h rt t hh k

y y k rs s kh k

+ += ⇒ = + + + += ⇒ = + +

Finally, observe that22

2 2 2 22 2

29( ) ( ) 1 rt s h kh k

+ = + + +

But since ( , )h k lies on 2,C we have 2 2 24 .h k r+ =



Thus,

2 2 2 299( ) (4 ) 94

t s r r + = = 2 2 2t s r⇒ + =

implying that ( , )G t s lies on 1.C

Since now you are in a good position to compare the two approaches, which one could you haverather chosen for this question, the pure-geometric one or the co-ordinate one!

Consider the circle 2 2 2.x y a+ = A chord of this circle is bisected at the point 1 1( , ).P x y What is the equation ofthis chord ?

Solution : Convince yourself that such a chord will be unique, since it must be perpendicular to the line joining theorigin to 1 1( , ),P x y as is clear from the figure below:

O (0,0)

AB

P x y( )1 1

Fig - 25

The chord is bisected at

ABP x y

OP AB.

( , ). Thus, we must have

1 1

⊥

The slope of AB then becomes 1

1

xy−

so its equation can be written simply as

11 1

1

( )xy y x xy

− = − −

2 21 1 1 1xx yy x y⇒ + = +

To make this equation look �better�, we subtract 2a from both sides

2 2 2 21 1 1 1xx yy a x y a+ − = + −

so that it can now be written concisely as

1 1 1 1( , ) ( , )T x y S x y=

Of course, this is easily generalised to the case when the equation of the circle is in the general form2 2 2 2 0;x y gx fy c+ + + + = the result obtained is the same. The next example discusses a good

application of this concept.

Example – 25



Find the locus of the mid-point of the chords of the circle 2 2 2x y a+ = which subtend a right angle at the centre.Solution:

Fig - 26

A

O ,(0 0)

y

M( )h,k B

x

AB

M h kAB

M

is a chord of the circlewhich subtends a right angle at the centre.

( , ) is the mid-pointof . We need to find the locus of .

Since AB is bisected at ( , ),M h k we can use the result obtained in the last example to write theequation of AB:

( , ) ( , )T h k S h k=2 2 2 2hx ky a h k a⇒ + − = + −2 2hx ky h k⇒ + = +

We can view the chord AB as a line intersecting the curve (circle) 2 2 2.x y a+ = Thus, we can obtainthe joint equation of OA and OB by homogenizing the equation of the circle using the equation of thechord AB:

22 2 2

2 2 0hx kyx y ah k+ + − = +

Joint equation ofOA and OB:

2 2 2 2 2 2 2( ) ( ) ( ) 0h k x y a hx ky⇒ + + − + = ...(1)Since OA and OB are perpendicular, we must have

2 2Coeff .of Coeff. of 0x y+ = in (1)2 2 2 2 2 2 22( ) 0h k a h a k⇒ + − − =

22 2

2ah k⇒ + =

Using ( , )x y instead of ( , ),h k we obtain the following equation as the locus of M:

22 2

2ax y+ =

This is a circle concentric with the original circle as might have been expected. Try solving this question using pure geometric considerations; the solution will be much simpler.

Example – 26



Consider two circles with the following equations:2 2

1 : 2 2 1 0C x y x y+ − − + =2 2

2 : 16 2 61 0C x y x y+ − − + =Find the values that a can take so that the variable line 2y x a= + lies between these two circles without touchingor intersecting either of them.

Solution : Observe carefully that what is variable about the variable line 2y x a= + is not its slope but itsy-intercept a. Thus, we can always adjust a so that line stays between the two circles. The followingdiagram makes this clear.

Fig - 27

y

x

The lines can vary in this range so that they staybetween the two circles.Thus as long on

< <

the line = 2 + staysbetween the two circles

a a ay x a

min max

(1,1) (8,1)

C1

(0, )amax

C2

(0, )amin

Evaluate maxa : The line max2y x a= + is a tangent to 1C if the perpendicular distance of the

centre (1, 1) of 1C from this line is equal to '1C s radius which is 1. Thus :

max2 11

5a− +

=

max1 5a⇒ + = ±

max 5 1a⇒ = − − max

since from the figure we can see that is definitelynegative

a

Example – 27



Evaluate mina : The distance of '2C s center (8, 1) from min2y x a= + must be equal to its radius

which is equal to 2. Thus :min16 1

25

a− +=

min15 2 5a⇒ + = ±

min 2 5 15a⇒ = − min

min 2

we have selected the larger of the twovalues possible since that is what corresponds to , i.e. because theline 2 lies .

ay x a above C

= + Thus, we obtain the allowed values of a as

2 5 15 5 1a− < < − −

A circle touches the line y x= at a point P such that 4 2OP = where O is the origin. The circle contains the

point (�10, 2) in its interior and the length of its chord on the line 0x y+ = is 6 2. Determine the equation of thecircle.Solution: As always, before starting with the solution, it is a good to draw a diagram of the situation described

to get a feel of it. Also, as far as possible, we should try to use pure-geometric considerations to cutdown on the (complicated) algebraic manipulations that would result otherwise.

Fig - 28

y

x

It should be more orless apparent that to be ableto contain the point (-10, 2)inside it, the centre ofthe circle must lie somewherein the shaded region.The point is then (- 4, - 4)since = 4 2Assume the centre to be at

( , ). The radius of this circle is then given by = where = ( + 4) + ( + 4)

POP

X h kr XP

r h k2 2 2

6 2B

A

P(-4,-4)

O

x + y = 0

y = x

X

' We have,( )PX y x⊥ =

4 14

kh+⇒ = −+ ...(1)

8 0h k⇒ + + = ...(2)

Example – 28



Also,

{ } 2 2 2 2Perpendicular dist.of from 0 BX AX AB r ABX x y = = − = −+ =

2 2 2( 4) ( 4) (3 2)2

h kh k

+⇒ = + + + − ...(3)

Using (1) and (2) in (3), we obtain2 2(4 2) 2( 4) 18h= + −

4 5h⇒ + = ±9,1h⇒ = −

Given the region in which X lies, h must be �9. Thus, from (2), k is 1 and the radius r is 5 2 .The required equation is therefore

2 2 2( 9) ( 1) (5 2)x y+ + − =2 2 18 2 32 0x y x y⇒ + + − + = ...(4)

For students used to rigor, it can finally be verified that the circle given by (4) does indeed contain thepoint (-10, 2) and thus our initial assumption of the region in which the centre X lies, was correct.

Let 1C and 2C be two circles with 2C lying inside 1.C A circle C lying inside 1C touches 1C internally and 2Cexternally. Determine the locus of the centre of C.

Solution: First of all, you must note that 2C is not concentric with 1.C All that is said is that 2C lies somewhere

inside 1.C

Let us first discuss what all would be involved in solving this question through a co-ordinate approach.

We could first assume a co-ordinate axes, say, with the x-axis lying along the line joining the centres of

1C and 2C and the origin at the centre of 1 :C

C1

O(0,0)C2

x(a, 0)

y

The centre of can be assumed to be ( , 0) as shown.The equations of and can then be written as

Ca

C C

2

1 2

C x + y r1 12 : = 2 2

C x – a y r22 : ( ) + = 2 2

2

Fig - 29

Example – 29



We could then assume the equation of C to be 2 2 2 2 0x y gx fy c+ + + + = and impose the necessaryconstraints on g, f and c so that C touches 2C externally and 1C internally. Recall the appropriate

constraints for two circles 1S and 2S touching externally and internally.

r2

O O r r1 2 1 2 = +

Fig - 30

r1

O2

(Circles touch externally)O O r r1 2 1 2 =| � |

(Circles touch internally)

O1

r1

O2

r2

O1

Imposing these constraints gives us the necessary conditions that g, f and c must satisfy and hence thelocus of the centre of C which is ( , ).g f− −However, we will be much letter off in using a pure-geometry approach here also as you�ll soon see.Assume an arbitrary circle C of radius r inside 1C with centre X, and which satisfies the given constraint:

Fig - 31

C

O2

B

O1

A

C1

The circle touches internally at and externally at . is the centre

of while, and respectively. The radii of

, and are , and respectively.

C CA C B X

C OC

C C C r r r

1

2

1

2

1 2 1 2

OC

2

1

are centres ofand C2

X

By the properties of circles touching internally and externally, we have1 1 1O X O A AX r r= − = −

2 2 2O X O B BX r r= + = +

1 2 1 2O X O X r r⇒ + = + ...(1)

(1) simply states the centre of C i.e. X, moves in such a way so that the sum of its distances from 1Oand 2O is constant. Thus, X must lie on an ellipse with 1O and 2O as the two foci !To sketch the path that X can take, we can follow the approach described in the unit on Complexnumbers. Fix two pegs at 1O and 2O and tie a string of length 1 2r r+ between these two pegs. Useyour pencil and the taut string as a guide to trace out the ellipse. This is the path on which the centre ofC can move.



Through an arbitrary fixed point 1 1( , ),P x y a variable line is drawn intersecting the circle 2 2 2x y a+ = at A and Brespectively. Tangents drawn to this circle at A and B intersect at Q. Find the locus of Q.

Solution :

Fig - 32

A

B

Q h k( )1

The tangents at and to the circle + = intersect at

. We wish to determinethe locus of whoseco-ordinates we have assumed to be ( )

A Bx y a

QQ

h k

2 2 2

1

P x ,y( )1 1

We can view this situation from a different perspective. We have a point ( , )Q h k from which we draw

tangents QA and QB to the circle 2 2 2.x y a+ = Thus, the equation of AB (which is the chord ofcontact) is

( , ) 0T h k =

2hx ky a⇒ + =

Now, this chord of contact also passes through 1 1( , )P x y so that

21 1hx ky a+ = ...(1)

What we have in (1) is a linear equation involving the variables h and k. Note that 1x and 1y areconstant. Thus, we can infer from (1) that ( , )h k lies on the line

21 1xx yy a+ = ...(2)

This is the required equation !

The line obtained in (2) is referred to as the polar of the point P with respect to the given circle. P isitself referred to as the pole of the polar.

Notice that the equation of the polar can be written concisely as

1 1 1 1( , ) 0 : Equation of the polar for the pole ( , )T x y P x y=

This example should show you that sometimes enormous simplifications are achieved using aco-ordinate geometrical approach rather than a pure-geometrical one. Co-ordinate geometry is notall that bad!

Example – 30 POLE AND POLAR



Q. 1 Three circle with radii 1 2,r r and 3r touch each other externally. The tangents at their points of contactmeet at a point whose distance from any point of contact is 4. Show geometrically that

1 2 3

1 2 3

16r r rr r r

=+ +

Q. 2 What is the equation of the tangent to 2 2 30 6 109 0x y x y+ − + + = at 1(4 1)?−

Q. 3 Find the tangents to 2 2 6 4 12 0x y x y+ − + − = parallel to 4 3 5 0.x y+ + =

Q. 4 Find the tangents to 2 2 2x y a+ = which make a triangle of area 2a with the axes.

Q. 5 If ( , )P a b and ( , )Q b a are two points ( ),b a≠ find the equation of the circle touching OP and OQ atP and Q where O is the origin .

Q. 6 Find the equation of the normal to 2 23 3 4 6 0x y x y+ − − = at (0, 0)

Q. 7 The line 2 1 0x y− + = is a tangent to a circle at (2, 5); its centre lies on 9.x y+ = Find its equation.

Q. 8 If 3 0x y+ = is a tangent to a circle whose centre is (2, �1), find the other tangent to the circle fromthe origin.

Q. 9 Find the locus of the point of intersection of tangents to the circle2 2 2 24 6 9sin 13cos 0x y x x+ + − + α + α = which are inclined at an angle of 2α to each other.

Q. 10 Prove that the intercept of the pair of tangents from the origin to 2 2 22 2 0,x y gx fy k+ + + + = on the

line y = λ is 2 2

2 kk gλ− .

TRY YOURSELF - II



Consider two circles 1C and 2C which intersect in two distinct points A and B. Our purpose is to determine the

equation of AB which will be termed the common chord of 1C and 2.C

Fig - 33

A

B

C2

C1AB

C CC

SS

is the common chord of the circles and . The equations of and C are given to be = 0 and

= 0 respectively.

1 2

1 2

1

2

Let the equations of the circles be2 2

1 1 1 12 2 0S x y g x f y c≡ + + + + =2 2

2 2 2 22 2 0S x y g x f y c≡ + + + + =

Consider the equation 1 2 0,S S S≡ − = i.e.

1 2 1 2 1 2(2 2 ) (2 2 ) 0S g g x f f y c c≡ − + − + − = ...(1)

Clearly, S is the equation of a straight line. Observe carefully that since A and B lie on both 1C and 2C , the co-

ordinates of A and B satisfy both 1S and 2S and thus 1 2 0.S S S≡ − = This means that 1 2 0S S− = represents astraight line passing through A and B. Thus, this is precisely the common chord !

1 2 0 : Equation of the common chordS S S≡ − =

You are urged to observe one important fact about S. It is perpendicular to the line joining the centres of 1C and

2;C this is but expected.

Fig - 34

A

C2

C1

Slope of O1 O2 is

P

B

O1 O2

( )–g , 2 –f2( )–g , 1 –f1

2 11

2 1m

f fg g

−= −

Slope of AB is from (1)

2 12

2 1mf fg g

−=

−

Thus, 1m 2m = � 1

The perpendicularity of 1 2O O and AB can be proved geometrically too in a straightforward manner.

Section - 3 EQUATIONS DESCRIBING CIRCLES



The length of the common chord can be easily evaluated using the Pythagoras theorem:2 2 2 2

1 1 2 22 2AB O A O P O A O P= − = −

where 1O P and 2O P are the perpendicular distances of 1O and 2O from the common chord respectively.

Notice an interesting fact : if the length of the common chord AB is 0, it is actually the common tangent to the twocircles 1C and 2C which will touch each other externally or internally.

Fig - 35

C2C1

PO1 O2

C2

C1

O2O1 P

If the two circles touch each other externally or internally (at ), then the common chord � = 0 actually represents the common tangents to the two circles at

PS S P1 2

Now an interesting question arises. Suppose that C1 and C2 lie external to each other and do not intersect. Whatdoes S1 � S2 = 0 represent in that case?

Fig - 36

C2

C1

S1= 0S2= 0

What does � = 0 represent in this case?Note that no common chord or common tangent can exist.

S S1 2

1 2 0S S− = is obviously a straight line. But it is obviously not a common chord or a common tangent since thesedo not exist in this case.

The answer is provided by a slight algebraic manipulation. Let a point 1 1( , )P x y be such that it satisfies 1 2 0.S S− =Thus,

1 1 1 2 2 1( , ) ( , )S x y S x y=

1 1 1 2 1 1( , ) ( , )S x y S x y⇒ =1 2

Assuming both sides are +vewhich is true if is external to and .

PC C



What does this equation tell us? P is a point such that the lengths of the tangents drawn from it to the two circles areequal. Thus, any point lying on the straight line 1 2 0S S− = will posess the property that the tangents drawn fromit to the two circles are equal. This line is termed the radical axis of the two circles.

Fig - 37

B

C1

For any point on the radical axis, tangents drawn from it to and are of equal length, i.e. =

p

C CPA PB

1 2

A

C2

This line is the radical axis

P

It should be obvious that in case of intersecting (or touching) circles, the common chord (or the common tangent)is itself the radical axis. For a situation as in Fig - 37 above, the radical axis exists but no common chord exists, Fora circle lying inside another circle, neither the radical axis nor the common chord exist :

Fig - 38

C2C1

The radical axis is the same as the common chord.

C2C1

X

P

The radical axis is the same as the common tangent.This should otherwise be obvious also since for any point on this line, the length of tangents to both and is

PC C PX1 2

C1

The radical axis is the same as the common tangent which is again abvious

No common chord or radical axis exist



Show that the radical axis of three circles, whose centres are non-collinear, taken two a + a time are concurrent.

Solution: The significance of the phrase �non-collinear� for the three centres should be clear to you : if thecentres are all collinear, the three radical axis will become parallel to each other instead of intersecting.

We assume the equations to the three circles to be 1 20, 0S S= = and 3 0.S = The three radical axisare therefore

1 1 2: 0R S S− =

2 2 3: 0R S S− =

3 3 1: 0R S S− =Observe that

1 2 3(1) (1) (1) 0R R R+ + =

which implies that 1 2,R R and 3R are concurrent. The point of concurrency is called the radical centreof the three circles :

R1

R2

R3

C3

C2

C1 X

R R R

X

X

1 2 3, and are the three radical axis, while , the point of concurrency is termed the radical centre.Tangents drawn from

to the three circleswill be of equal lengths

Fig - 39

____________________________________________________________________________________

Before proceeding we must discuss some properties of two intersecting circles; in particular, we need to understandwhat we mean by the angle of intersection of two circles.

Consider two intersecting circles 1C and 2C with radii 1r and 2r respectively and centres at 1O and 2O respectively:

C2C1

C CA B

1 2 and intersect at and .

A

B

Fig - 40

O2O1

Example – 31



The angle of intersection of the two circles can be defined as the angle between the tangents to the two circles attheir point(s) of intersection, which will be the same as the angle between the two radii at the point(s) ofintersection. In particular, for example, 1C and 2C in the figure above intersect at an angle 1 2.O AO∠

The most important case we need to consider pertaining to intersecting circles is orthogonal circles, meaning thatthe angle of intersection of the two circles is a right angle. In that case, the 1 2.O AO∠ above will become a right -angled one so that

2 2 21 2 1 2O A O A O O+ = ...(1)

If the two circles and 1C and 2C have the equations

2 2 2 21 1 1 1 1 1 1 1: 2 2 0 ;S x y g x f y c r g f c+ + + + = = + −

2 2 2 22 2 2 2 2 2 2 2: 2 2 0 ;S x y g x f y c r g f c+ + + + = = + −

then the condition (1) becomes

2 2 21 2 1 2r r O O+ =

2 2 2 2 2 21 1 1 2 2 2 1 2 1 2( ) ( )g f c g f c g g f f⇒ + − + + − = − + −

1 2 1 2 1 22( )g g f f c c⇒ + = +

Thus, the condition that the two circles given by 1S and 2S must satisfy in order to be orthogonal is

1 2 1 2 1 22( ) : Orthogonal circlesg g f f c c+ = +

As an exercise, verify that the following pairs of circles intersect orthogonally :

(a) 2 21

2 22

: 6 8 24 0: 8 6 24 0

S x y x yS x y x y

+ − − + =+ − − + =

(b) 2 21

2 22

: 5 3 7 0: 8 6 18 0

S x y x yS x y x y

+ + + + =+ − + − =

As another example, suppose that we have to find the angle at which2 2

12 2

2

: 6 4 11 0: 4 6 9 0

S x y x yS x y x y

+ − + + =+ − + + =

intersect.

We have,

1 2 1 22; 2; (3, 2); (2, 3)r r O O= = ≡ − ≡ −



Referring to Fig - 40, let the angle of intersection, 1 2O AO∠ be θ . By the cosine rule in 1 2 ,O AO∆ we have

2 2 21 2 1 2

1 2

1cos2 2

r r O Or r

θ + −= =

4πθ⇒ =

Thus, the two circles intersect at an angle equal to .4π

A circle C passes through (1, 1) and cuts the following two circles orthogonally :

2 21

2 22

: 8 2 16 0: 4 4 1 0

S x y x yS x y x y

+ − − + =+ − − − =

Find the equation of C.

Solution: We assume the equation of C to be

2 2: 2 2 0S x y gx fy c+ + + + =

Applying the condition of orthogonality of S with 1S and 2,S we obtain :

with 1S : 2( 4 ) 16g f c− − = +

8 2 16g f c⇒ + + = − ...(1)

with 2S : 2( 2 2 ) 1g f c− − = −

4 4 1g f c⇒ + + = ...(2)

The third condition can be obtained using the fact that C passes through (1, 1):

2 21 1 2 (1) 2 (1) 0g f c+ + + + =

2 2 2g f c⇒ + + = − ...(3)

Solving (1), (2) and (3), we obtain

7 23, , 53 6

g f c= − = = −

Thus, the equation of the circle C is2 2: 3 3 14 23 15 0S x y x y+ − + − =

Example – 32



Prove that the locus of the centres of the circles cutting two given circles orthogonally is their radical axis.

Solution: This assertion means that any circle cutting two given circles orthogonally will have its centre lying onthe radical axis of the two given circles.

Assume the two fixed given circles to have the following equations:

2 21 1 1 1: 2 2 0S x y g x f y c+ + + + =

2 22 2 2 2: 2 2 0S x y g x f y c+ + + + =

Let the variable circle be 2 2: 2 2 0S x y gx fy c+ + + + = so that its centre is ( , )g f− − whose locuswe wish to determine.

Applying the condition for orthogonality, we obtain

1 1 12( )gg ff c c+ = + ...(1)

2 2 22( )gg ff c c+ = + ...(2)

By (1) � (2), we obtain

1 2 1 2 1 22 ( ) 2 ( )g g g f f f c c− + − = −

Using ( , )x y instead of ( , ),g f− − we obtain the locus as 1 2 1 2 1 22 ( ) 2 ( ) ( ) 0x g g y f f c c− + − + − =

which is the same as the equation of the radical axis, i.e. 1 2 0.S S− =

We can also prove this assertion of this question in a very straightforward manner using a pure geometricapproach. Let C be a circle intersecting the two given circles 1C and 2C orthogonally as show in thefigure below:

Fig - 41

C1

C CC

OA O A

OB O B

intersects and orthogonally.

Thus, and

1

2

1

2

⊥

⊥A

C2

O

O2O1

B

C

1OA O A⊥ and 2OB O B⊥ implies that OA and OB are simply the tangents drawn from O to 1C and

2C . Since OA and OB are the radii of the same circle C, we have .OA OB=

Example – 33



Thus, O is a point such that tangents drawn from it to the two circles 1C and 2C are equal in length,

which implies that O lies on the radical axis of 1C and 2C .

From this property, a straightforward corollary follows. For three fixed circles, a circle with centre atthe radical centre and radius equal to the length of the tangent from it to any of the circles will intersectall the three circles orthogonally. As an exercise, find the equation to the circle C cutting the followingthree circles orthogonally :

2 21 : 2 3 7 0C x y x y+ − + − =

2 22 : 5 5 9 0C x y x y+ + − + =

2 23 : 7 9 29 0C x y x y+ + − + =

Prove that two circles, both of which pass through the point (0, )a and (0, )a− and touch the line ,y mx c= + will

cut orthogonally if 2 2 2(2 )c a m= +

Solution: It should be clear that the y-axis is the common chord of the two circles 1C and 2C :

C2

C1

y C C

C Cx

- axis is the commonchord of and Also, the centres of both and with lie on the - axis

1 2

1 2

Fig - 42

y

(0 )1,a

(0 ),-a

x

Let the centres of 1C and 2C be 1( ,0)g− and 2( ,0)g− so that their radii become 2 21g a+ and

2 22g a+ respectively. Their equations then become :

2 2 21 1: 2 0C x y g x a+ + − =

2 2 22 2: 2 0C x y g x a+ + − =

Example – 34



The line y mx c= + touches both 1C and 2C so that the perpendicular distance of the centres of 1C

and 2C from this line must be respectively equal to their radii. Thus, we obtain

1 2 2121

mg cg a

m

−= +

+ and 2 2 2

221mg c

g am−

= ++

2 2 2 21 12 (1 ) 0g mcg a m c⇒ + + + − =

and 2 2 2 22 22 (1 ) 0g mcg a m c+ + + − =

Thus, 1g and 2g are the roots of the equation

2 2 2 22 (1 ) 0g mcg a m c+ + + − =

so that

2 2 21 2 (1 )g g a m c= + −

Finally, 1C and 2C are orthogonal if the condition

1 2 1 2 1 22( )g g f f c c+ = +

is satisfied, i.e.

2 2 2 22{( (1 ) ) (0)(0)} 2a m c a+ − + = −

2 2 2(2 )c a m⇒ = +

This is the required relation for 1C and 2C to be orthogonal.

Consider the following two circles:

2 21

2 22

: 16 0: 8 12 16 0

S x yS x y x y

+ − =+ − − + =

Circles are drawn which are orthogonal to both these circles. Tangents are drawn from the centre of the variablecircle to 1S . Find the mid locus of the mid-point of the chord of contact so formed.

Solution: The centre of the variable circle, say 1 1( , ),x y lies on the radical axis of the two given circles, i.e. on

1 2 0S S− = by the result obtained in Example - 33.

Example – 35



Thus, 1 1( , )x y must satisfy

1 2 0S S− = or 8 12 32 0x y+ − =

1 18 12 32 0x y⇒ + − = ...(1)

From 1 1( , ),x y two tangents are drawn to 1.S The equation of the chord of contact is therefore

1 1( , ) 0T x y =

1 1 16xx yy⇒ + = ...(2)

Let the mid-point of the chord of contact so formed be ( , ).M h k We can also write the equation ofthe same chord of contact using the equation we derived for a chord bisected at a given point.

Thus, the chord of contact can also be represented by the equation

( , ) ( , )T h k S h k=

2 2hx ky h k⇒ + = + ...(3)

Since (2) and (3) are the same lines, we have

1 12 2

16x yh k h k= =

+

1 12 2 2 2

16 16;h kx yh k h k

⇒ = =+ + ...(4)

Using (4) in (1), we finally obtain a relation in ( , )h k :

1 18 12 32 0x y+ − =

2 2 2 2

128 192 32h kh k h k

⇒ + =+ +

2 2 4 6 0h k h k⇒ + − − =

Using ( , )x y instead of ( , )h k we obtain the required locus as

2 2 4 6 0x y x y+ − − =



Q. 1 Find the equation of the circle which intersects 2 2 6 4 3 0x y x y+ − + − = orthogonally, passes through(3, 0) and touches the y-axis.

Q. 2 At what angle do the circles

2 21

2 22

: 4 6 11 0: 2 8 13 0

S x y x yS x y x y

+ − + + =+ − + + =

intersect ?

Q. 3 Find the co-ordinates of the point from which the tangents drawn to the following three circles are ofequal lengths :

2 21

2 22

2 23

: 3 3 4 6 1 0: 2 2 3 2 4 0: 2 2 1 0

S x y x yS x y x yS x y x y

+ + − − =+ − − − =+ − + − =

Q. 4 Find the locus of the centre of the circle passing through ( , )a b and orthogonal to 2 2 2.x y k+ =

Q. 5 Let 1 1( , )A a b and 2 2( , )B a b be two fixed points and O be the origin. Circles are drawn on OA and

OB as diameters. Prove that the length of the common chord is 1 2 2 1a b a bAB−

.

TRY YOURSELF - III



By a family of circles, we will mean a set of circles satisfying some given property (or properties). For example, thefamily of circles with each circle having its centre lying in the first quadrant and touching both the co-ordinate axescan be represented by the equation

2 2 2( ) ( )x a y a a− + − = ...(1)

where a is a positive real number.

The important point to observe is that a is a variable here. As we vary a, we get different circles belonging to thisfamily, but due to the constraint imposed by (1), all circles of this family satisfy the specified property.

y

x

Some members of the family of circles given by (1).The centre of a circle is ( , ) and itsradius is , where is a real positive variable. As is varied, we obtain different members in this family

a aa a

a

Fig - 43

We intend to discuss in this section certain families that are of significant importance. In all cases, the family will berepresented by an equation containing a real variable, which when varied will give rise to different members of thisfamily.

TYPE 1:FAMILY OF CIRCLES PASSING THROUGH THE INTERSECTION POINTS OF A GIVEN CIRCLE AND A GIVEN LINE

We are given a fixed circle and a fixed line with equations 0S = and 0L = respectively. We want tofind out the equation of the family of the circles passing through the points of intersection of 0S = and

0L = .

The circles = 0 and = 0 are fixed. The dotted circles represent some of the members of the family of circles passing through the intersection points

and of = 0 and = 0

S L

A B S L

L = 0

BS = 0

A

Fig - 44

Section - 4 FAMILY OF CIRCLES



By now, it should be apparent to you how to specify the equation of this family in terms of a realvariable. Any circle F belonging to this family can be written as

0F S Lλ≡ + = where λ ∈ ! .

The truth of this assertion can be easily verified. F as defined from this equation is definitely a circlesince it satisfies the required constraints for its equation to be a circle. To see this, let

2 2: 2 2 0S x y gx fy c+ + + + =

: 0L px qy r+ + =

We then have

: 0F S Lλ+ =

2 2 (2 ) (2 ) 0x y g p x f q y c r⇒ + + + + + + + =λ λ λ

which is definitely the equation of a circle. To see that F passes through A and B, note that since A andB satisfy both the equations 0S = and 0,L = they will obviously also satisfy the equation 0.S Lλ+ =

Thus, 0F S Lλ≡ + = is the required family of circles. As we vary λ , we will get different circlesbelonging to this family. In particular, note that for 0,λ = we get the circle 0S = itself, while in thelimit ,λ → ∞ we get the �circle� 0,L = which is actually a line but can be considered a circle withcentre lying at infinity and an infinitely large radius.

Write the equation for

(a) the family of circles passing through 1 1( , )x y and 2 2( , )x y

(b) the family of circles touching the line 0L = at 1 1( , )x y

Solution (a) Till now we have seen how to write the equation for a family of circles passing through the intersectionpoints of a circle and a line. Thus, in the current example, we first define a fixed circle passingthrough 1 1( , )x y and 2 2( , )x y and the line through these two points.

The fixed circle S can easily be taken to be the one with 1 1( , )x y and 2 2( , )x y as the end points ofa diameter :

1 2 1 2: ( )( ) ( )( ) 0S x x x x y y y y− − + − − =

The fixed line can be written using the two-point form :

1 11 1

2 1 2 12 2

1: : 1 0

1

x yy y x xL L x yy y x x

x y

− −= ⇒ =− −

Example – 36



Given two fixed points ( , ) and ( , ), we define the fixed circleas the one with these pointsbeing the end-points of a diameter. The fixed line is simply the line passingthrough these two points.

x y x y1 1 2 2

S = 0

L = 0( , )x y2 2( , )x y1 1

Fig - 45

We can now write the required family of circles as :

: 0F S Lλ+ =

1 2 2 2( )( ) ( )( ) 0x x x x y y y y L⇒ − − + − − + =λ ...(1)

where 0L = represents the line through the two given points

(b) We now want the family of circles to be such that each circle touches 0L = at 1 1( , )x y .

We could evaluate it by letting 2 1x x→ and 2 1y y→ in (1). In this limit, 0L = will simplybecome the tangent to any member of the family F given by (1). (Convince yourself about thispoint.

Thus, the required family is

2 21 1: ( ) ( ) 0F x x y y Lλ′ − + − + =

TYPE 2: FAMILY OF CIRCLES TOUCHING A GIVEN CIRCLE AT A GIVEN POINT

Let the equation of the fixed circle be

2 2: 2 2 0S x y gx fy c+ + + + =

and let there be a point 1 1( , )P x y lying on this circle. We wish to determine the equation of the familyof circles touching S at P.

The dotted circles aresome of the members of the family of circles in which each circle touches at S P

S = 0

P( , )x y1 1

Fig - 46



We can write the equation of the tangent to 0S = at P as

1 1 1 1: ( ) ( ) 0T xx yy g x x f y y c+ + + + + + =

Once we have a circle ( 0)S = and a line ( 0)T = intersecting or touching the circle, we can write theequation of the family of circles passing through the point (s) of intersection of the circle and the line,using the result derived in the last article. Thus, the required family can be represent as

: 0F S Tλ+ =

2 21 1 1 1: 2 2 ( ( ) ( ) 0F x y gx fy c xx yy g x x f y y cλ⇒ + + + + + + + + + + + =

2 21 1 1 1: (2 ) (2 ) 0F x y g x g x f y f y c gx fy c⇒ + + + + + + + + + + + =λ λ λ λ λ λ λ

As we vary ,λ we will obtain different members belonging to this family.

TYPE 3:FAMILY OF CIRCLES PASSING THROUGH THE INTERSECTION POINT(S) OF TWO GIVEN CIRCLES

Let the two fixed circles be

2 21 1 1 1: 2 2 0S x y g x f y c+ + + + =

2 22 2 2 2: 2 2 0S x y g x f y c+ + + + =

and their points of intersection be 1 1( , )A x y and 2 2( , ).B x y In case the two circles touch each other,A and B will be the same.

We wish to determine the family of circles passing through A and B.

The dotted circles represent some members of the family of circles in which each member passes through and A B

Fig - 47

S = 2 0 B

S = 1 0

A

You might be able to extrapolate from the last few cases that the equation representing this familywill be

1 2: 0F S Sλ+ =



You can verify this by writing the equation for F in standard form and observing that it does indeedrepresent a circle. Also, since (the co-ordinates of) A and B satisfy both 1 0S = and 2 0,S = theyhave to satisfy the equation for F. Note one important point: λ cannot be equal to �1 otherwise F willbecome the common chord of 1 0S = and 2 0S = instead of representing a circle.

Find the equation of the circle which passes through the points of intersection of the circles

2 21 : 6 2 4 0S x y x y+ − + + =

2 22 : 2 4 6 0S x y x y+ + − − =

and whose centre lies on the line y = x.

Solution: Let the required equation be S = 0. Then, by the previous article, we can find some λ ∈ ! and1λ ≠ − such that

1 2 0S S S≡ + λ =

2 2(1 ) (1 ) (2 6) (2 4 ) 4 6 0S x y x y⇒ ≡ + λ + + λ + λ − + − λ + − λ = ...(1)

The centre of S from this equation comes out to be

( 3) (1 2 )centre ,1 1λ − − λ ≡ − − + λ + λ

Since the centre lies on the line y = x, we have

( 3) (1 2 )1 1λ − − λ− = −+ λ + λ

43

⇒ λ =

We now substitute this value back in (1) to obtain the equation for S:

2 27 7 10 10: 4 03 3 3 3

S x y x y+ − − − =

2 2: 10 10 12 0S x y x y⇒ + − − − =

A family of circles passing through the points A (3, 7) and B (6, 5) cuts the circle 2 2 4 6 3 0.x y x y+ − − − = Showthat the common chord of the fixed circle and the variable circle(belonging to the family) will always pass througha fixed point. Find that point.

Example – 37

Example – 38



Solution: We can write the equation to the specified family of circles by first writing the equation L of the line AB:

7 2:3 3

yLx− =− −

: 2 3 27L x y⇒ + =

The required family can now be written as

: ( 3)( 6) ( 7)( 5) 0F x x y y L− − + − − + λ = where λ ∈ !

2 2: (2 9) (3 12) (53 27 ) 0F x y x y⇒ + + λ − + λ − + − λ =

The common chord of F and the given circle S is :

0S F− =

(5 2 ) (6 3 ) (27 56) 0x y⇒ − λ + − λ + λ − =

(5 6 56) (2 3 27) 0x y x y⇒ + − − λ + − =

1 2 0L L⇒ + µ =

That the common chord can be written like 1 2 0L L+µ = implies that it will always pass through theintersection point of L1 and L2, what ever the value of µ may be. This intersection point can be

obtained (by simultaneously solving L1 and L2) to be 232,3

.

Find the equation of the circle which touches the line 0x y− = at the origin and bisects the circumference of the

circle 2 2 2 3 0x y y+ + − = .

Solution: In example -36, we evaluated the equation of the family of circles all touching a given line at a givenpoint. Here, the given line is 0x y− = and the given point is (0, 0). Thus, the equation of the family is

2 2: ( 0) ( 0) ( ) 0F x y x y− + − + λ − =

2 2: 0F x y x y⇒ + + λ −λ = ...(1)

We need to find the value of λ for which the circle in (1) bisects the circumference of the given circle2 2: 2 3 0,S x y y+ + − = which means that the common chord of the required circle and S will the

diameter of S.

Example – 39



A

BThe variable circle

S = 0

If the variable circlebisects the circumference of , this mean that the common chord must be the diameterof .

S

ABS

Fig - 48

The common chord AB is

0F S− =

( 2) 3 0x y⇒ λ − λ + + =

Since this is the diameter of S, the centre of S, i.e. (0, �1), must lie on it (satisfy its equation). Thus, weobtain λ as

(0) ( 2)( 1) 3 0λ − λ + − + =

5⇒ λ = −

Finally, we substitute this value of λ back in (1) to get the required equation of the circle as

2 2 5 5 0.x y x y+ − + =

Two circles, each of radius 5 units, touch each other at (1, 2). If the equation of their common tangent is4 3 10,x y+ = find the equation of the circles.

Solution: The following diagram explains the situation better:

4 + 3 = 10x y

The two circlestouch eachother at (1, 2).

Fig - 49

C2C1

S2 = 0

(1, 2)S1 = 0

We describe here two alternatives that can be used to solve this question

Example – 40



Alternative - 1: Find the centres of the two circles

The slope of the common tangent is 4 .3

− Therefore, the slope of C1C2 is 3 ,4 i.e.

3tan4

m = θ =

3sin5

⇒ θ = and 4cos5

θ =

Using the polar form, we can write the co-ordinates of any point on the line C1C2:

1 2cos sinx y r− −= =θ θ

1 cos , 2 sinx r y r⇒ = + θ = + θ

Substituting 5r = ± gives the two centres as should be apparent from the figure.Thus, the two centres are

1 (5, 5)C ≡ and 2 ( 3, 1)C ≡ − −

Thus, the two equations are2 2

1 : 10 10 25 0S x y x y+ − − + =2 2

2 : 6 2 15 0S x y x y+ + + − =

Alternative - 2: Use a family of circles approache

The family of circles touching L = 0 at 1 1( , )x y can be written, as describedearlier, as

2 21 1( ) ( ) 0x x y y L− + − + λ =

In the current case, this becomes2 2( 1) ( 2) (4 3 10) 0x y x y− + − + λ + − =

2 2 (4 2) (3 4) (5 10 ) 0x y x y⇒ + + λ − + λ − + − λ = ...(1)

The radius of the required circle is 5. Thus,2

2 23 4(2 1) (5 10 ) 52λ − λ − + − − λ =

This is a quadratic in λ which gives two values of λ :

2λ = ±

Using these values in (1), we obtained the two required circles.



Q. 1 Find the equations of the circles with radius 4 and passing through the points of intersection of

2 21 : 2 4 4 0S x y x y+ − − − =

2 22 : 10 12 40 0S x y x y+ − − + =

Q. 2 Find the equation of the circle whose diameter is the common chord of the circles 2 2 2( )x a y a− + =

and 2 2 2( )x y b b+ − = .

Q. 3 Find the equation of the circle passing through (2, 1) and touching the line 2 1x y+ = at (3, �1).

Q. 4 Tangents are drawn from the origin to 2 2 6 4 12 0.x y x y+ + + − = Find the equation of the circlepassing through the points of contact of these tangents and the origin.

Q. 5 Find the equation of a circle which touches the line 5x y+ = at the points (�2, 7) and cuts the circle2 2 4 6 9 0x y x y+ + − + = orthogonally.

TRY YOURSELF - IV



The circle 2 2 4 4 4 0x y x y+ − − + = is inscribed in a triangle which has two of its sides along the co-ordinateaxes. If the locus of the circumcentre of the triangle is

2 2 0x y xy k x y+ − + + =

find the value of k.

Solution : The situation is described clearly in the figure below:

(2,2)

y

x

The circle = 0is fixed. The line is variable,intersecting the axis in ( , 0)and (0, ) respectively.We are concerned with the locusof the circumcentre of .

S

A aB b

OAB∆

Fig - 50

B(0 ), b

S= 0

0 ( 0)a, A

L

The equation of L is, using the intercept form,

: 1x yLa b+ =

The distance of the centre of S, i.e. (2, 2) from L must equal the radius of S which is 2. Thus,

2 2

2 2 12

1 1a b

a b

+ −=

+

We now use the fact that (2, 2) is negative since

(2, 2) and the origin lie on the same side of and

(0, 0) is negative

L

LL

2 22 2 2 0a b ab a b⇒ + − + + = ...(1)

From pure geometric considerations, the circumcentre C of OAB∆ lies on AB and is in fact, themid-point of AB.

Thus,

,2 2a bC ≡

SOLVED EXAMPLES

Example – 01



Slightly manipulating (1), we obtains

2 2

02 2 2 2 2 2a b a b a b + − + + =

...(2)

The locus of ,2 2a b

is given by (2). Using ( , )x y instead of , ,2 2a b

we obtain

2 2 0x y xy x y+ − + + = ...(3)

Upon comparing (3) with the locus specified in the question, we obtain 1.k =

Consider a curve 2 22 1ax hxy by+ + = and a point P not on the curve. A line drawn from P intersects the curve at

points Q and R. If the product PQ PR⋅ is independent of the slope of the line, then show that the curve is a circle.

Solution : Since distances are involved from a fixed point, it would be a good idea to use the polar form of theline to write the co-ordinates of Q and R.

Let P be 1 1( , )x y and let θ denote the slope of the variable line. For any point ( , )x y lying on this lineat a distance r from P, we have

1 1

cos sinx x y y r

θ θ− −= =

1 1cos ; sinx x r y y rθ θ⇒ = + = +

If ( , )x y lies on the given curve, it must satisfy the equation of the curve :

2 21 1 1 1( cos ) ( sin ) 2 ( cos )( sin ) 1a x r b y r h x r y rθ θ θ θ+ + + + + + =

2 2 21 1 1{ cos sin sin 2 } 2{ cos sin ( cosa b h r ax by h xθ θ θ θ θ θ⇒ + + + + +

2 21 1 1 1 1sin )} 2 1 0y r ax by hx yθ+ + + + − = ...(1)

This quadratic has two roots in r, say 1r and 2r , which will actually correspond to PQ and PRsince Q and R lie on the curve. Thus, PQ PR⋅ being independent of θ means that 1 2r r for (1) isindependent of θ i.e.

2 21 1 1 12 2

2 1cos θ sin θ sin 2θax by hx y

a b h+ + −+ +

is independent of θ

2 21 1 1 12 1

cos 2θ sin 2θ2 2

ax by hx ya b a b h

+ + −⇒+ − + +


Example – 02



2 21 cos 2 1 cos 2The denominator was obtained in this form by writing cos as and sin as 2 2

+ −

θ θθ θ

2 21 1 1 1

22

2 1

{sin(2 }2 2

ax by hx y

a b a b h θ φ

+ + −⇒+ − + + +


The denominator was obtained by combining the two trignometric terms. tan is 2

a bh

−

φ

⇒ This is only possible when 2

2 02

a b h− + =

⇒ a b= and 0h =

Thus, the equation of the given curve reduces to

2 2 1ax ay+ =

which is clearly the equation of a circle.

The equation of a circle is : 2 ( ) (2 ) 0S x x a y y b− + − = where , 0.a b ≠ Find the condition on a and b if two

chords each bisected by the x-axis, can be drawn to the circle from the point , .2bP a

Solution: Observe that the centre of S is ,2 4a b

and P lies on the circle:

Fig - 51

The centre of the circle is ( ); S

y

xthe point lies on the circle.In fact, is a diameter of the circle

POP

a2

b4

,a2

b4

,( )

a b2

,( )

O

P

Example – 03



Let us consider a chord PQ of the circle bisected at the x-axis, say, at the point ( ,0).h We can writethe equation of PQ as :

( ,0) ( ,0)T h S h=

where ( , )S x y is 2 2 02bx y ax y+ − − =

2( ) ( 0)2 4a bhx x h y h ah⇒ − + − + = −

2 02 4 2a b ahh x y h ⇒ − − + − =

...(1)

Since this chord passes through , ,2bP a

the co-ordinates of (P) must satisfy (1). Thus,

2 02 4 2 2a b b ahh a h − − + − =

2 223 0

2 2 8ah a b h⇒ − − − =

2 22 3 0

2 2 8ah a bh⇒ − + + = ...(2)

We want to have two such possible chords PQ bisected at the x-axis. Thus, we must have twodistinct values of h which can happen if the discriminant of (2) is positive. Thus,

2 2 29 44 2 8a a b > +

2 22a b⇒ >

This is the required condition that a and b must satisfy.

Let 1T and 2T be two tangents drawn from (�2, 0) to the circle 2 2 1.x y+ = Determine the circles touching C and

having 1T and 2T as their pair of tangents. Also find all the possible pair wise common tangents to these circles.

Example – 04



Solution: Note that two such circles can be drawn, as shown in the figure below :

Ry

x

C C

x x

1 2

1 2

and are the twopossible circles, with centres ( , 0) and ( , 0)−

Fig - 52

B x ,( 0)2

QP

S( 2 0)- ,

x + y = 12 2 ( )1

Centre at,0x−

C1

OA

R'

C2

Let the radii of 1C and 2C be 1r and 2r respectively. We can evaluate the coordinates of both A andB very easily from geometrical considerations. We have

ASP OSQ BSR∆ ∆ ∆∼ ∼

AS OS BSAP OQ BR

⇒ = =

Now, 1 1 2 22 , , 2, 1, 2 ,AS x AP r OS OQ BS x BR r= − = = = = + =

1 2

1 2

2 2 21

x xr r− +⇒ = =

Also, note that 1 11x r= + and 2 21 .x r= + Thus,

1 2

1 2

2 221 1x x

x x− +⇒ = =− −

143

x⇒ = and 2 4x =

113

r⇒ = and 2 3r =



Thus, the equations to 1C and 2C are

2 22

14 1:3 3

C x y + + = 2 2 2

2 : ( 4) 3C x y− + =

Now we determine all the possible common tangents to 1C and 2C .

Two such possible tangents are simply SR and SR′ which both touch 1C and 2C on the same side.There will be two other possible common tangents, each having A and B on the opposite side of it asshown in the figure below :

y

x

These tangents aretermed thetransverse tangentsto and .C C1 2

Fig - 53

PA

QC2

Bθ

C1

X

Let X have the co-ordinates ( ,0).x Note that .APX BQX∆ ∆∼Thus,

AX BXAP BQ

=

443

1/ 3 3

x x+ −⇒ =

45

x⇒ = −

Also,1 3 5sin

4 5 4 3 8APAX

θ = = =− +

2 2

5 5tan398 5

θ⇒ = =−

Thus, the two transverse tangents pass through 4 ,05−

and have slopes 5 .39

± Their equations

are therefore5 4

539y x = ± +



Find the values of a for which the line 0x y+ = bisects two chords drawn from a point 1 2 1 2,2 2

a aP + −

to

the circle 2 2 1 2 1 2: 02 2

a aS x y x y + −+ − − =

.

Solution: This question is somewhat similar to Example -3. Note that the point P lies on S.

Since the mid-point M of the chord(s) drawn from P lies on 0,x y+ = we can assume its co-ordinatesto be ( , ).h h− The equation of the chord of S bisected at M then becomes:

( , ) ( , )T h h S h h− = −

2( ) ( ) 22 2

hx hy x h y h h h hλ λ λ λ⇒ − − + − − = − + ...(1)

where the substitution 1 2 1 2,2 2

a aλ λ+ −= = has been done for convenience.

Since the chord (1) passes through ( , ),P λ λ its co-ordinates must satisfy (1). Thus,

2( ) ( ) 22 2

h h h h h h hλ λλ λ λ λ λ λ− − + − − = − +

2 2 24 3( ) ( ) 0h hλ λ λ λ⇒ − − + + = ...(2)

Since we want two possible chords to exist satisfying the given property, the quadratic in (2) mustyield distinct values of h. Thus, its discriminant must be positive :

2 2 29( ) 16( )λ λ λ λ− > +

Substituting the values of λ and λ and simplifying, we obtain

2 4 0a − >

( , 2) (2, )a⇒ ∈ −∞ − ∪ ∞

These are the required values of a.

If 1 0S = and 2 0S = are circles having radii 1r and 2r respectively, prove that the circles 1 2

1 2

: 0S SSr r+ = and

1 2

1 2

' : 0S SSr r− = intersect orthogonally.

Example – 05

Example – 06



Solution: Let the equation of the two given circles be2 2 2 2 2

1 1 1 1 1 1 1 1: 2 2 0 ;S x y g x f y c r g f c+ + + + = = + −

2 2 2 2 22 2 2 2 2 2 2 2: 2 2 0 ;S x y g x f y c r g f c+ + + + = = + −

Then, the equation of S and S ' are

2 2 1 2 1 2 1 2

1 2 1 2 1 2 1 2

1 1: ( ) 2 2 0g g f f c cS x y x yr r r r r r r r

+ + + + + + + + =

2 2 1 2 1 2 1 2

1 2 1 2 1 2 1 2

1 1: ( ) 2 2 0g g f f c cS x y x yr r r r r r r r ′ + − + − + − + − =

We now check the orthogonality condition for S and S ', i.e. we check whether

' ' '2( )S S S S S Sg g f f c c+ = +

Thus,

1 2 1 2 1 2 1 2

1 2 1 2 1 2 1 2' '

1 2 1 2 1 2 1 2

2( ) 2 21 1 1 1 1 1 1 1S S S S

g g g g f f f fr r r r r r r r

g g f f

r r r r r r r r

+ − + −

+ = +

+ − + −

2 2 2 21 1 2 2

2 21 2

2 21 2

21 1

g f g fr r

r r

+ +− =

−

1 22 2

1 2

2 21 2

21 1

c cr r

r r

−

=

−

...(1)

whereas

1 2 1 2

1 2 1 2'

1 2 1 2

1 1 1 1S S

c c c cr r r rc c

r r r r

+ −+ = +

+ −

1 22 2

1 2

2 21 2

21 1

c cr r

r r

−

=

−

... (2)

The equality of (1) and (2) implies that the two circles are orthogonal.



Let 1 0S = and 2 0S = be two circles with radii 1r and 2r respectively. Find the locus of the point P at which the

two circles subtend equal angles.

Solution: Consider 1 0S = which subtends an angle α (say) at ( , )P h k

( )-g,-f C1

P h k( , )

S1= 0

T

Fig - 54

We have,

1 1C T r= and 1( , )PT S h k=

Thus, in 1 ,C PT∆ we have

1

1

tan2 ( , )

rS h k

=α...(1)

We can write an exactly analogous equation for 2S which subtends the same angle α at P:

2

2

tan2 ( , )

rS h k

α = ...(2)

From (1) and (2), we have

1 2

1 2( , ) ( , )r r

S h k S h k=

Using ( , )x y instead of ( , ),h k we obtain the required locus as

1 22 2

1 2

( , ) ( , )S x y S x yr r

=

Circles are drawn passing through the origin O to intersect the co-ordinate axes at points P and Q such thatm OP n OQ⋅ + ⋅ is a constant. Show that the circles pass through a fixed point other than the origin.

Example – 07

Example – 08



Solution: Consider one such circle as shown in the figure below:

Fig - 55

y

x

Given that + = (a constant)

we need to show that will always pass through

a fixed point

mOP nOQ k

S

O

C

S

q(0, )Q

P p,( 0)

We have OP p= and .OQ q= Thus,

mp nq k+ = ...(1)

Now, observe that the centre C of the circle will be ,2 2p q

and its radius will be OC. Thus, the

equation of the circle is

2 2 2 2

2 2 4 4p q p qx y − + − = +

2 2 0x y px qy⇒ + − − = ...(2)

From (1), we can find q in terms of p and substitute in (2) so that (2) becomes

2 2 0mp kx y px yn− + − + =

2 2{ ( ) } { } 0n x y ky p nx my⇒ + − + − + =

We see that the variable circle can be written as

0 0 0,S pL p+ = ∈ !

where 2 20 ( ) 0S n x y ky≡ + − = and 0 0L nx my≡ − + = . Thus, the variable circle will always pass

through the two intersection points of S0 and L0, one of them obviously being the origin. The otherfixed point can be obtained by simultaneously solving the equations of 0S and 0L to be

2 2 2 2,mk nkm n m n + +



The two curves

2 21 : 2 2 2 0S ax hxy by gx fy c+ + − − + =

2 22 : 2 ( ) 2 2 0S a x hxy a a b y g x f y c′ ′ ′ ′− + + − − − + =

intersect at four concyclic points , ,A B C and .D Let P be the point , .g g f fa a a a′ ′+ +

′ ′+ + Find the value

of 2 2 2

2 .PA PB PCPD

+ +

Solution: From what we�ve learnt in this chapter, we can write the equation of an arbitrary curve S passingthrough the intersection points of two given curves, here 1S and 2S , as

1 2 0S S Sλ≡ + = where λ ∈ !

Here,

2 2: ( ) 2 (1 ) ( ( )) 2 ( )S a a x h xy b a a b y x g g′ ′ ′+ + − + + + − − +λ λ λ λ 2 ( ) (1 ) 0y f f c λ′− + + + = ..(1)

Since this curve S passes through the four concyclic points , , ,A B C D its equation in (1) must bethat of a circle. Thus, we impose the necessary constraints :

2 2Coeff. of Coeff. of ( )x y a a b a a b′ ′= ⇒ + = + + −λ λ

1λ⇒ =

Coeff. of 0 1xy = ⇒ =λ

Thus, for 1,λ = the equation in (1) represents a circle. This equation now becomes :

2 2: ( ) ( ) 2( ) 2( ) 2 0S a a x a a y g g x f f y c′ ′ ′ ′+ + + − + − + + =

The centre of S therefore is , .g g f fa a a a

′ ′+ + ′ ′+ +

But this is the same as the point P!. Thus, P is the

centre of the circle passing through , , ,A B C D so that

PA PB PC PD= = =

Thus, we simply have

2 2 2

2 3PA PB PCPD

+ + =

Example – 09



Let x, y and z be the lengths of the perpendiculars dropped from the circumcentre in a triangle ABC to the sidesBC, CA, and AB respectively. The lengths of BC, CA and AB are a, b and c respectively. Show that

4a b c abcx y z xyz+ + =

Solution: We can attempt this question through a co-ordinate approach but the nature of the problem posedsuggests that a plane geometric approach would be better since no equations need to be determinedhere.

Refer to the following diagram which shows the situation clearly :

The circumcentre is .All lengths have been marked.Note that = 2 = 2

=

O

COB AA

θ⇒ θ

Fig - 56

Ba

cb O

Px

θzy

A

C

In ,OPB∆ we have

tan θ tan2aAx

= ∠ = ...(1)

Similarly, tan2bBy

∠ = and tan2CCz

∠ = ...(2)

Now, the angles ,A B∠ ∠ and C∠ are those of a triangle so that

A B C π∠ + ∠ + ∠ =

tan( ) tan( )A B Cπ⇒ ∠ + ∠ = − ∠

tan C= − ∠

tan tan tan1 tan tan

A B CA B

∠ + ∠⇒ = − ∠− ∠ ∠

tan tan tan tan tan tanA B C A B C⇒ ∠ + ∠ + ∠ = ∠ ∠ ∠

Using (1) and (2) in the relation above, we obtain the required relation.

Example – 10



Let ( ,0)A a be a point on the circle 2 2 2.x y a+ = Through another point (0, ),B b chords are drawn to meet thecircle at points Q and R. Find the locus of the centroid of .AQR∆

Solution : The following diagram shows an example of the situation described :

The points and are fixed. The line is of variable slope. We need to determine the locus of the centroid of

A BBQR

AQR∆

Fig - 57

Q

R

A a, ( 0)

B , b(0 )

The equation of BQR can be written as y mx b= + where m is variable. The intersection points ofBQR with the circle (Q and R) are given by simultaneously solving the system of equations

2 2 2x y a+ =y mx b= +

Thus,2 2 2( )x mx b a+ + =

2 2 2 2(1 ) 2 0m x mbx b a⇒ + + + − = ...(1)

If we assume the co-ordinates of Q and R to 1 1( , )x y and 2 2( ),x y we have from (1)

1 2 2

21

mbx xm

+ = −+

Thus,2

1 2 1 2 2 2

2 2( ) 2 21 1

m b by y m x x b bm m

+ = + + = − + =+ +

Let the centroid of AQR∆ be ( , ).h k We have

1 2 1 20,3 3

a x x y yh k+ + + += =

1 2 1 23 , 3x x h a y y k⇒ + = − + =

2 2

2 23 , 31 1

mb bh a km m

⇒ − = − =+ +

These two relations can be used to eliminate m and obtain a relation in ( , ).h k To specify the locusconventionally, we use ( , )x y instead of ( , ).h k The equation obtained is (verify) :

22 2 2 2 0

3 3 9ax y ax by+ − − + =

Example – 11



Find the equation of the family of circles touching the lines given by2 2 2 1 0x y y− + − =

Solution: The equation to the pair of lines can be factorised to yield the individual lines as

1 0x y− + = and 1 0.x y+ − =

From the following diagram, observe carefully that for a circle to touch both the lines above, its centremust be on one of their angle bisectors because only then will the distance of the centre from the twolines be equal.

x + y - = 1 0

y

x

(0,1) ( 1)λ,

(0, )λ

x - y + = 1 0

For a circle tobe able to touch both the lines,its centre must lieon one of their angle bisectors

Fig - 58

The angle bisectors can be seen from inspection to be 0x = and 1.y = Thus, the centre of thevariable circle can be assumed to be:

(i) (0, )λλ ∈ !

: The distance of (0, )λ from the two lines is 12

λ −

Thus, the equation of the variable circle in this case is2

2 2 ( 1)( )2

x y λλ −+ − = ...(1)

(ii) ( ,1)λλ ∈ !

: The distance of ( ,1)λ from the two lines in | |2λ

Example – 12



Thus, the equation of the variable circle in this case is

22 2( ) ( 1)

2x y λλ− + − = ...(2)

The equations (1) and (2) represent the required family of circles.

A circle of radius r passes through the origin O and cuts the axes at A and B. Show that the locus of the foot of theperpendicular from O to AB is

2 2 2 22 2

1 1( ) 4x y rx y

+ + =

Solution: Let the co-ordinates of A and B be ( ,0)a and (0, )b respectively, so that (like in Example - 8), theequation to the variable circle becomes

2 2 0x y ax by+ − − =

A a,( 0)

P h,k( )

B b(0, )

s

O

The equation for is + - - = 0

Note that since = , is a diameter of

the circle.

Sx y ax by

AOBAB

2 2

Fig - 59

π2

We have,

2 2 24a b r+ = ...(1)

Let the foot of perpendicular P have the co-ordinates ( , ).h k Since ,OP AB⊥ we obtain

1k bh a× = −−

2 2 2 2

2 2 2k h h k h ka b ra b

+ +⇒ = = =+

2 2 2 2

2 2,rk rha bh k h k

⇒ = =+ +

...(2)

Using (2) in (1) and ( , )x y instead of ( , ),h k we obtain the required locus.

Example – 13



Find the condition so that the chord cos sinx y p+ =α α subtends a right angle at the centre of the circle2 2 2x y a+ = .

Solution:

x + y = a2 2 2

y

x

Fig - 60

x + y = pcos sin ααB

O

A

We can obtain the joint equation J to the pair of lines OA and OB by homogenizing the equation of thecircle using the equation of the chord AB.

22 2 2 cos sin: x yJ x y a

p ++ =

α α... (1)

For J to represent two perpendicular lines, we must have in (1),

2 2Coeff. of oeff. of 0x C y+ =

which upon simplification yields the required condition as

2 22a p=

Suppose that the lines 1 1 1 0a x b y c+ + = and 2 2 2 0a x b y c+ + = intersect the co-ordinate axes at points ,A B and

,C D respectively. Find the condition that must be satisfied if these four points are to be concyclic.

Example – 14

Example – 15



Solution: The co-ordinate of ,A B and ,C D can be evaluated to be 1 1 2

1 1 2

,0 , 0, , ,0c c ca b a

− − −

and

2

2

0, .cb

−

Fig - 61

The four points , and , are given to be concyclic

A B C D

A COD

B

L a x + b y + c = 2 2 2 2 : 0

L a x + b y + c = 1 1 1 1 : 0

y

x

Instead of resorting to a detailed calculation, we simply use the result on tangents and secants thatwe�ve already derived earlier:

2OA OC OD OB l⋅ = ⋅ =

where l is the length of the tangent drawn from O to the circle. This gives

1 2 2 1

1 2 2 1

c c c ca a b b

− ⋅ − = − ⋅ −

1 2 1 2 0a a b b⇒ − =

This is the required condition !



ASSIGNMENTASSIGNMENTASSIGNMENTASSIGNMENTASSIGNMENT

[ LEVEL - I ]1. Find the equation of the circle passing through (�2, 1) and tangent to 3 2 6 0x y− − = at (4, 3).2. If two circles cut a third circle orthogonally, prove that their radical axis will pass through the centre of the

third circle.3. Find the point from which the tangents drawn to the following three circles are of equal lengths :

2 21 : 16 16 0S x y x+ − + =

2 22 : 3 3 36 8 0S x y x+ − + =

2 23 : 16 12 8 0S x y x y+ − − + =

4. Prove that the length of the common chord of the two circles 2 2 2( ) ( )x a y b c− + − = and

2 2 2( ) ( )x b y a c− + − = is 2 24 2( )c a b− −5. Prove that the locus of the centre of the variable circle which touches externally the

2 2 6 6 14 0x y x y+ − − + = and the y-axis is 2 10 6 14 0.y x y− − + =6. Find the equation of the circle whose radius is 3 and which touches internally the circle

2 2 4 6 12 0x y x y+ − − − = at (�1, �1).

7. Prove that the circle 2 2 2 0x y ax c+ + + = and 2 2 2 0x y by c+ + + = will touch one another if

2 2

1 1 1a b c+ = .

8. If the circles 2 2 10 16 0x y x+ − + = and 2 2 2x y r+ = intersect in real and distinct points, prove that2 < r < 8.

9. Find the point at which 2 2 4 2 4 0x y x y+ − − − = and 2 2 12 8 12 0x y x y+ − − − = touch each other.

10. Find the locus of the mid-point of a variable chord of the circle 2 2 2 2 2x y x y+ − − − which subtends an

angle 23π

at the centre.

11. Find the locus of the mid point of the variable chord of the circle 2 2 2x y a+ = which subtends a right angleat (c, 0)

12. If the squares of the lengths of the tangents from a point P to the circles 2 2 2 2 2 2,x y a x y b+ = + = and2 2 2x y c+ = are in A.P., show that a2, b2 and c2 are in A.P.

13. P is a variable point on y = 4. Tangents from P touch 2 2 4x y+ = at A and B. The parallelogram PAQB iscompleted. What is the locus of Q?

14. The angle between T1 and T2 is ,3π

where T1 and T2 are tangents to 2 2 4x y+ = and 2 2 9x y+ =

respectively. Find the locus of the intersection of T1 and T2 .15. Show that tangents from (�1, 7) to 2 2 25x y+ = are at right angles to each other.



[ LEVEL - II ]

16. Find the circumcircle of the triangle formed by the three lines

1 : 6 0L x y+ − =

2 : 2 4 0L x y+ − =

3 : 2 5 0L x y+ − =17. The following two circles intersect in A and B:

2 2 21 : 2 0S x y ax c+ + − =

2 2 22 : 2 0S x y bx c+ + − =

A line through A meets one circle at P while a parallel line through B meets the other circle at Q. Show thatthe locus of the mid-point of PQ is a circle.

18. A variable circle passes through the fixed point A (a, b) and touches the y-axis. Show that the locus of theother end of the diameter through A is

2( ) 4y b ax− =

19. Suppose that the point P(h, 0) divides a chord AB of the circle 2 2 2x y a+ = with slope tan ,θ in the ratio

:1.m Prove that2 2 22 cos

1h h a

m mθ − = +

20. Find the equation of the circle, the equation of two of whose normal is 2 3 6 2 0x x y xy+ + + = and which

is large enough to just contain the circle 2 2 4 3 .x y x y+ = +21. Find the possible values of α for which the point ( 1, 1)α − α + lies in the larger segment of the circle

2 2 6 0x y x y+ − − − = made by the chord 2x y+ =22. Prove that the square of the tangent that can be drawn from any point on one circle to another circle is equal

to twice the product of the perpendicular distance of the point from the radical axis of the two circles andthe distance between their centres.

23. Find the radius of the smallest circle circumscribing the following three circles :2 2

1 : 4 5 0S x y y+ − − =2 2

2 : 12 4 31 0S x y x y+ + + + =2 2

3 : 6 12 36 0S x y x y+ + + + =

24. A variable chord through a fixed point Q (t, 0 )meets a fixed circle 2 2 2: ,S x y a+ = at A and B. Find thelocus of a point P on this chord such that QA, QP and QB are in (a) A. P. (b) G.P.

25. If H is the orthocentre of a triangle ABC, prove that the radii of the circumcircles of the triangles ABC andHBC are of the same length.



ANSWERS

TRY YOURSELF -I

1. [15,5]

2. ( )2

2 2 0x y mx yc

λ+ + − =

3. ( )2 23 34

g f c

+ −

4.2 2

2 2

8 2 7 03 7 12 0

x y x yx y x y + − − + =

+ − − + =

5.2 21 22 485

17 17 289x y

+ + − =

6. ( ) ( )2 22 1 1x y − + + =

7. ( )2 213 2 13 0x y y + − − =

8. 2 2 0x y x y + − − =

10.9 ,2

2 −



TRY YOURSELF -II

2. [ ]11 2 46 0x y− − =

3. [ ]4 3 19 0; 4 3 31 0x y x y+ + = + − =

4. [ ]x y a± ± =

5. ( ) ( )2 2

2 2 2 22 0a bx y x y a ba b

++ − + + + = +

6. [3x � 2y] =0

7. ( ) ( )2 26 3 20x y − + − =

8. [ ]3 0x y− =

9. 2 2 4 6 9 0x y x y + + − + =

TRY YOURSELF -III

1. 2 2 6 6 9 0x y x y + − − + =

2. [ ]/ 4π

3.16 31,21 63

− −

4. 2 2 22 2ax by a b k + = + +

TRY YOURSELF -IV

1.2 2

2 2

2 2 18 22 69 02 15 0

x y x yx y y + − − + =

+ − − =

2. ( )( ) ( )2 2 2 2 2x y a b ab bx ay + + = +

3. ( )2 23 23 4 35 0x y x y + − = + =

4. 2 2 3 2 0x y x y + + + =

5. 2 2 7 11 38 0x y x y + + − + =



ASSIGNMENT

ANSWER

LEVEL - I

1. ( )2 27 4 82 55 0x y x y + + − + =

3.10 2,3 3

−

6. 2 25 5 8 14 32 0x y x y + − − − =

9.2 4,5 5

− −

10. 2 2 2 2 1 0x y x y + − − + =

11. ( )2 2 2 22 2 0x y cx c a + − + − =

13. ( )( )2 2 22y x y x y = + +

14. ( )2 23 76x y + =

LEVEL II

16. 2 2 17 19 50 0x y x y + − − + =

17. ( )2 2 0x y a b x + + + =

20. 2 2 6 3 45 0x y x y + + − − =

21. ( )1,1−

23. [approx 7.3]

24.( )( )

2 2

2 2 22 0

a x y tx

b x y tx a

+ =

+ − + =

circles

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