# Free Objective Test 01 Practice Test - 11th and 12th

The centre of the circle passing through (0,0) and (1,0) and touching the circle x2+y2=9 can be

A.

(12,12)

B.

(12,2)

C.

(32,12)

D.

(12,32)

#### SOLUTION

Solution : B

Let the centre of the circle be (x, y)

x=0+12

=12

r=32

x2+y2=32

y=±2

So there are two circles possible. The length of common chord of the circles (xa)2+y2 =a2 and x2+(yb)2 = b2 is

A.

2a2+b2

B.

aba2+b2

C.

2aba2+b2

D.

2aba2b2

#### SOLUTION

Solution : C

Equation of common chord is axby = 0

Now length of common chord

= 2r21p21 = 2r22p22

Where r1 and r2 are radii of given circles and p1,p2 are the perpendicular distances from centres of circles to common chords.

Hence required length = 2a2a4a2+b2 = 2aba2+b2.

The power of (2,1) with respect to the circle 2x2+2y28x6y+k = 0 is positive if

A.

0<k<12

B.

-12<k<12

C.

k>12

D.

k<12

#### SOLUTION

Solution : C

2x2+2y28x6y+k = 0

x2+y24x3y+(k2) = 0

4+183+k2>0

k26>0k>12

Equation of circle touching the line |x-2|+|y-3|=4 will be

A.

(x2)2+(y3)2=12

B.

(x2)2+(y3)2=4

C.

(x2)2+(y3)2=10

D.

(x2)2+(y3)2=8

#### SOLUTION

Solution : D

Perpendicular distance from centre to tangent = radius r=|2+39|2=42=422=22 Equation of circle is (x2)2+(y3)2=8

If the line x + 2by + 7 = 0 is diameter of the circle x2+y26x+2y=0, then b =

A. 3
B. -5
C. -1
D. 5

#### SOLUTION

Solution : D

Here the centre of circle (3, -1) must lie on the line x + 2by + 7 = 0
Therefore, 3 - 2b + 7 = 0 b = 5

A circle with centre at the origin and radius equal to a meets the axis of x and A and B. P(α) and Q(β) are two points on this circle so that αβ=2γ, where γ is a constant. The locus of the point of intersection of AP and BQ is

A.

x2y22aytanγ=a2

B.

x2+y22aytanγ=a2

C.

x2+y2+2aytanγ=a2

D.

x2y2+2aytanγ=a2

#### SOLUTION

Solution : B Coordinates of A are (–a, 0) and of P are (acosα,asinα)
Equation of AP is y=asinαa(cosα+1)(x+a)y=tan(α2)(x+a)(1)
Similarly equation of BQ = y=asinβa(cosβ1)(xa)y=cot(β2)(xa)(2)
From (1) and (2), tan(α2)=yα+x,tan(β2)=axy
Now αβ=2γ
tanγ=tan(α/2)tan(β/2)(a+x)y+(ax)y=yα+xaxy1+yα+x.axytanγ=tan(α/2)tan(β/2)1+tan(α/2)tan(β/2)=x2+y2a22ayx2+y22aytanγ=a2 which is required locus.

Two vertices of an equilateral triangle are (–1, 0) and (1, 0) and its third vertex lies above the x-axis, the equation of the circumcircle is

A. 3x2+3y223y=3
B. 2x2+2y232y=3
C. x2+y22y=1
D. None

#### SOLUTION

Solution : A

Let A(–1, 0), B(1, 0) and C(0, b) be the vertices of the triangle, since C lies on the locus of points equidistance from A(–1, 0), and B(1, 0) i.e., y-axis.
Then
AB=AC
1+b2=2
b2=3
b=3  [ b>0]
Since the triangle is equilateral, the centre of the circumcircle is at the centroid of the triangle which is (0,13).
Thus the equation of the circumcenter is,
(x0)2+(y13)2=(10)2+(013)2
x2+y2(23)y+13=1+13
3x2+3y223y=3

The angle at which the circles (x1)2+y2= 10 and x2+(y2)2=5 intersect is

A.

π6

B.

π4

C.

π3

D.

π2

#### SOLUTION

Solution : B

r1=10r2=5d=(1)2+(2)2=5cosθ=r21+r22d22r1r2=10+552105=102×5×2=12 Consider a family of circles which are passing through the point (–1, 1) and are tangent to x-axis. (h, k) are the co-ordinates of the centre of the circles, then the set of values of k is given by the interval

A. 12k12
B. k12
C. < k < 12
D. k12

#### SOLUTION

Solution : D

Since, circles are passing through the point (–1, 1) and touching X-axis, the equationof circles can be written as -
(h+1)2+(K1)2=K2
h2+2h+22K=0
If h is real the Δ0
4- 4.1. (2 – 2K)  0
= 4 - 8 + 8k 0
= 8k 4
k12

The number of common tangents to the circles x2+y2+2x+8y23=0 and x2+y24x10y+19=0 is

A.

1

B.

2

C.

3

D.

4

#### SOLUTION

Solution : C

To find the no. of common tangents, let's find the relation between distance between the centers of these circles and sum of their radii -
Co-ordiantes of the center of first circle is (-1, - 4)
Co-ordiantes of the center of second circle is (2, 5)
thus the distance between them is -  (32+92)
=  3(10)
Now, the sum their radii will be -
Radius of the first circle is -  2(10)
Radius of the second circle is - (10)
Sum of them will be -  3(10)
Since, the distance between the centres is equal to sum of their radii
Number of common tangents will be 3.

If the lengths of the tangents from two points A, B to a circle are 6, 7 respectively. If A, B are conjugate points then AB =

A. 5
B. 85
C. 852
D. none

#### SOLUTION

Solution : B

Let the equation of the circle be x2 + y2=a2 and A (x1,y1), B(x2,y2)
A, B are conjugate points x1x2+y1y2=a2 (1)
Length of the tangents from A, B are 6, 7 respectively x21+y21a2 = 6, x22+y22a2= 7
x21+y21a2 = 36
x22+y22a2 = 49
x21+y21+x21+y222a2 = 85
x21+y21+x21+y222
(x1 x2 + y1 y2) = 85 [From (1)]
(x21+x222x1x2) + (y21+y222y1y2) = 85 (x1x2)2 + (y1y2)2 = 85 AB2 = 85 AB = 85

The equation of the circle passing through the point (1, –2) and having its centre on the line 2x – y – 14 = 0 and touching the line 4x + 3y – 23 = 0

A. x2 + y2 + 8x + 12y + 27 = 0
B. x2 + y2 - 12y + 27 = 0
C. x2 + y2 - 8x + 12y - 27 = 0
D. x2 + y2 - 8x + 12y + 27 = 0

#### SOLUTION

Solution : D

Let P(1, -2) and the centre be C(a, b)
Centre lies on 2x - y - 14 = 0 2a - b - 14 = 0 b = 2a - 14 (1)
Circle touches the lines 4x + 3y - 23 = 0. Lets say that this point of touching is Q.
We know that,
CQ = CP = Radius of the circle.
(a1)2+(b+2)2 = |4a + 3b  23|42 + 32
(a1)2+(2a14+2)2 = |4a+3(2a14)23|5 [From (1)]
5a22a+1+4a248a+144 = |10a65| 55a250a+145 = 5|2a13|
5a2 - 50a + 145 = (2a - 13)2 = 4a2 - 52a + 169
a2 + 2a - 24 = 0 (a + 6)(a - 4) a = 4
Case 1 : a = 4 b = 2(4) - 14 = - 6. Centre C = (4, - 6)
Radius = CP = (41)2+(6+2)2 = 9+16 = 5
The circle equation is (x4)2+(y+6)2=52 x2 + y2 - 8x + 12y + 27 = 0
Case 2 : a = - 6 b = 2(- 6) - 14 = 26 Centre C = (-6, - 26)
Radius = CP = (61)2+(26+2)2 = 49+576 = 625 = 25
The circle equation is (x+6)2+(y+26)2=252 x2 + y2 + 12x + 52y + 87 = 0
The circles are x2 + y2 - 8x + 12y + 27 = 0, x2 + y2 + 12x + 52y + 87 = 0

If the chord y = mx + 1 of the circle x2+y2=1 subtends an angle of measure 45 at the major segment of the circle then value of m is

A. 2
B. -2
C. -1
D. none of these

#### SOLUTION

Solution : C

x2+y2(ymx)2=0
Coefficient x2 +Coefficient y2=0
m=+1 or -1

The equation of the image of the circle x2+y2 - 6x - 4y + 12 = 0 by the line mirror x + y - 1 = 0

A. x2+y2 + 2x + 4y + 4 = 0
B. x2+y2 - 2x + 4y + 4 = 0
C. x2+y2 + 2x + 4y - 4 = 0
D. x2+y2 + 2x - 4y - 4 = 0

#### SOLUTION

Solution : A

For the circle x2+y2 - 6x - 4y + 12 = 0.
This circle has centre C(3, 2) and radius (r) = 1
Image of (3, 2) w.r.t the line x + y - 1 = 0      = (-1, -2)
Equation of the circle with centre (-1, -2) and radius 1 is,
(x+1)2+(y+2)2 = 1
Image of x2+y2 - 6x - 4y + 12 = 0 w.r.t x + y - 1 = 0 is,
x2+y2 + 2x + 4y + 4 = 0

If x + y + k = 0 is a tangent to the circle x2+y2 - 2x - 4y + 3 = 0 then k =

A. ± 20
B. –1, – 5
C. ± 2
D. 4

#### SOLUTION

Solution : B

Radius = Distance from the centre C(1, 2) to x + y + k = 0
1+43 = |1 + 2 + k|2 |k+3| = 2 k + 3 = ± 2 k = -1, - 5