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11/25/13 Question Revision

Question 1.

The equation of the circle concentric w ith the circle x2+ y2- 8x + 6y - 5 = 0and passing through the point (-2, -7), is

[A] x2+ y2+ 8x + 6y - 27 = 0

[B] x2+ y2- 8x + 6y - 27 = 0

[C] x2+ y2+ 8x + 6y + 27= 0

[D] None of these

Your Answ er: --Correct Answer: B

Explanation:

The centre of the given circle is at (4,- 3). Therefore, the centre of the required circle is also at (4,- 3). Since the point (- 2, - 7) lies on

the circle, the distance of the centre from this point is the radius of the circle.

Hence, the equation of the circ le becomes

(x - 4)2+(y + 3)2= 52

or x2+ y2- 8x + 6y - 27 = 0.

Question 2.

The equation of the circle which touches the axis of yat a distance + 4from the origin and cuts off an intercept 6f rom the axis of xis

[A] x2+ y2- 10x - 8y + 16 = 0

[B] x2+ y2+ 10x - 8y + 16 = 0

[C] x2+ y2- 10x + 8y + 16 = 0

[D] None of these

Your Answ er: --

Correct Answer: D

Explanation:

Let the centre of the circle be C. Let the circle touches y-axis atAand cuts off intercept DEfrom x-axis. Let Bbe the mid point of DE.

Now , OA = 4 BCand BD =3.

Radius of the circle = =

Also, OB = AC= radius = 5

Centre of the circle is (5, 4)

Thus, equation of the required circle is

(x - 5)2+(y - 4)2=(5)2


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Each side of the square is landAB, ADare the coordinate axes

Coordinates of Band Dare (1, 0)and (0, 1)respectively.

Since BAD = 90 , BDis a diameter of the circumcircle

of squareABCD.

Equation of circumcircle is

(x - l)(x - 0) +(y - 0)(y - l) = 0

or x2+ y2= l(x + y).

Question 5.

The circlesx2+ y2- 10x + 16 = 0andx2+ y2= r2intersect each other in two distinct points if

[A] r < 2

[B] r > 8

[C] 2 < r < 8

[D] 2 r 8

Your Answ er: --

Correct Answer: C

Explanation:

Centres of the given circles are

C1 (5, 0) and C2 (0, 0)

Also, their radii are r1= 3and r2= r.

Since the two circles cut each other in tw o distinct points, therefore

| r1- r2| < C1C2< r1+ r2

r - 3 < 5 < r + 3 2 < r < 8.

Question 6.

If the equation of a circle is

(4a - 3)x2+ ay2+ 6x - 2y + 2 = 0, then its centre is

[A] (3,-1)

[B] (3, 1)

[C] (-3, 1)

[D] None of these

Your Answ er: --

Correct Answer: C


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Explanation:

Since the given equation represents a circle, therefore, 4a - 3 = a i.e. a = 1

( Coef f ic ients of x2and y2must be equal).

So, the circle becomes x2+ y2+ 6x - 2y + 2 = 0

The coordinates of centre are (- 3,1).

Question 7.

The equationx2+ y2- 8x + 6y + 25 = 0represents

[A] a circle

[B] a pair of straight lines

[C] a point

[D] None of these

Your Answ er: --

Correct Answer: C

Explanation:

The given equation can be w ritten as

(x - 4)2+(y + 3)2= 0

x = 4, y =- 3, w hich gives a point.

Question 8.

If the coordinates of tw o consecutive vertices of a regular hexagon w hich lies completely above the x-axis, are (-2, 0) and (2, 0), then

the equation of the circle, circumscribing the hexagon, is

[A]

[B]

[C]

[D]

Your Answ er: --

Correct Answer: A

Explanation:

It is clear from the figure that coordinates of the circumcentre are (0, 2 tan 600) or

Also, circumradius =


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Explanation:

Question 11.

If the circlex2+ y2+ 6x + 8y + a = 0bisects the c ircumference of the circlex2+ y2+ 2x - 6y - b = 0,then a + bis equal to

[A] 38

[B] - 38

[C] 42

[D] None of these

Your Answ er: --

Correct Answer: B

Explanation:

Given circles are

S1: x2+ y2+ 6x + 8y + a = 0 ...... (i)

and S1: x2+ y2+ 6x - 6y - b = 0 ..... (ii)

The equation of common chord of the tw o circles is

S1- S2= 0 i.e. 4x + 14t + (a + b) = 0 ........(iii)

Since the circle S1bisects the circumference of c ircle S2,therefore, (i) passes through the centre of c ircle S2 i.e. (- 1, 3)

4 (-1) + 14(3)+ a + b = 0 a + b = -38.

Question 12.

The lines 2x - 3y = 5and 3x - 4y = 7are diameters of a circle having area as 154sq. units. Then the equation of the c ircle is

[A] x2+ y2+ 2x - 2y = 62

[B] x2+ y2+ 2x - 2y = 47

[C] x2+ y2- 2x + 2y = 47

[D] x2+ y2- 2x + 2y = 62

Your Answ er: --

Correct Answer: C


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Explanation:

Centre of the circle is the intersection of the diameters. So, solving

Centre is (1,- 1)

Area of the circ le = r2

Equation of the circle is

(x - 1)2+(y + 1)2= 72

x2- 2x + 1 + y2+ 2y + 1 = 49

x2+ y2- 2x + 2y = 47.

Question 13.

The centre of the three circles

x2+ y2= 1, x2+ y2+ 6x - 2y = 1andx2+ y2- 12x + 4y = 1

[A] form a right angled triangle

[B] form an isosceles triangle

[C] are collinear

[D] None of these

Your Answ er: --

Correct Answer: C

Explanation:Centres of the three given circles areA(0, 0), B(- 3, 1) and C(6,- 2) respectively.

Area of ABC=

Hence, the centresA, B, Care collinear

Question 14.

The locus of the centres of the circles w hich cut the circlesx2+ y2+ 4x - 6y + 9 = 0

and x2+ y2- 4x + 6y + 4 = 0orthogonally is

[A] 8x - 12y + 5 = 0

[B] 8x + 12y - 5 = 0

[C] 12x - 8y + 5 = 0

[D] None of these

Your Answ er: --

Correct Answer: A

Explanation:

Let the equation of one of the circles be


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x2+ y2+ 2g x + 2f y + c = 0

Since it cuts the given circles orthogonally,

2g(2) + 2f(-3) = c + 9

and 2g(-2) + 2f(3) = c + 4

i.e. 4g - 6f = c + 9 and -4g + 6f = c + 4

On subtracting, w e get: 8g - 12f = 5

i.e. - 8(- g) + 12(- f) = 5

So the locus of (- g, - f)is -8x + 12y = 5.

Question 15.

The distance from the centre of the circle x2+ y2= 2x to straight line passing though the points of intersection of the tw o circles x2+ y2+

5x - 8y + 1 = 0 and x2+ y2- 3x + 7y - 25 = 0 is

[A]

[B] 2

[C] 3[D] 1

Your Answ er: --

Correct Answer: B

Explanation:

The equation of the straight line passing through the points of intersection of given circles is

(x2+ y2+ 5x - 8y + 1) - (x2+ y2- 3x + 7y - 25) = 0

i.e. 8x - 15y + 26 = 0 ...... (i)

Also, centre of the circlex2+ y2- 2x = 0is (1, 0)

Distance of the point (1, 0) from the straight line (i) is =

Question 16.

A square is inscr ibed in the circlex2+ y2- 2x + 4y + 3 = 0. Its sides are parallel to the coordinate axes. Then one vertex of the square is

[A]

[B]

[C][D] None of these

Your Answ er: --

Correct Answer: D

Explanation:

The centre of the given circle is (1,- 2). Since the sides of the square inscribed in the circle are parallel to the coordinate axes, so the

x-coordinate of any vertex cannot be equal to 1and its y - coordinate cannot be equal to - 2. Hence none of the points given in (a), (b)

and (c) can be the vertex of the square.

Question 17.


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If the equation of the incircle of an equilateral triangle isx2+ y2+ 4x - 6y + 4 = 0, then the equation of the c ircumcircle of the triangle is

[A] x2+ y2+ 4x + 6y - 23 = 0

[B] x2+ y2+ 4x - 6y - 23 = 0

[C] x2+ y2- 4x - 6y - 23 = 0

[D] None of these

Your Answ er: --

Correct Answer: B

Explanation:

Given equation of the incircle is

x2+ y2+ 4x - 6y + 4 = 0.

Its incentre is (- 2, 3) and inradius = .

Since in an equilateral triangle, the incentre and the circumcentre coincide,

Circumcentre = (- 2, 3).

Also, in an equilateral triangle, circumradius = 2(inradius)

Circumradius = 2 . 3 = 6

The equation of the circumcircle is

(x + 2)2+(y - 3)2=(6)2

or x2+ y2+ 4x - 6y - 23 = 0.

Question 18.

The PQRis inscribed in the circlex2+ y2= 25. If Qand Rhave coordinates (3, 4) and (- 4, 3) respectively, then QPRis equal

to

[A] /2

[B] /3

[C] /4

[D] /6

Your Answ er: --

Correct Answer: C

Explanation:

Let m1= slope of OQ= and m2= slope of OR= . As m1m2= -1, QOR = /2.

Thus, QPR=


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( angle subtended at the centre of a circle is double the angle subtended in the alternate segment).

Question 19.

The tangent to the c irclex2+ y2= 5at the point (1,-2) also touches the

circle x2+ y2- 8x + 6y + 20= 0.Then its point of contact is

[A] (3,-1)[B] (-3, 0)

[C] (-1, -1)

[D] (-2, 1)

Your Answ er: --

Correct Answer: A

Explanation:

Equation of tangent to the circ le

x2+ y2= 5at (1, - 2) is x -2y - 5 = 0 ...... (i)

Let this line touches the circle

x2+ y2- 8x + 6y + 20 = 0at (x1, y1)

Equation of tangent at (x1, y1) is

or x(x1- 4) + y(y1+ 3) - 4x1+ 3y1+ 20 = 0 ..... (ii)

Now (i) and (ii) represent the same line

- 2x1+ 8 = y1 or 2x1+ y1 - 5 = 0

Only the point (3,- 1) satisfies it. Hence the point of contact is (3,- 1)

Question 20.

For the tw o circlesx2+ y2= 16andx2+ y2- 2y = 0,there is/are

[A] one pair of common tangents

[B] tw o pairs of common tangents

[C] three common tangents

[D] no common tangent

Your Answ er: --

Correct Answer: D

Explanation:

Given circles are

S1: x2+ y2- 16 = 0 ......(i)

and S2: x2+ y2- 2y = 0 .......(ii)

Centre of S1is C1: (0, 0) and radius r1= 4

Centre of S2is C2: (0,1) and radius r2= 1


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Since | C1C2| < | r1- r2|, \ S2is completely w ithin S1and hence there are no common tangents to the tw o circles.

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