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

### Question 1

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

(12,12)

(12,−√2)

(32,12)

(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.

### Question 2

The length of common chord of the circles (x−a)2+y2 =a2 and x2+(y−b)2 = b2 is

2√a2+b2

ab√a2+b2

2ab√a2+b2

2ab√a2−b2

#### SOLUTION

Solution :C

Equation of common chord is ax−by = 0

Now length of common chord

= 2√r21−p21 = 2√r22−p22

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 = 2√a2−a4a2+b2 = 2ab√a2+b2.

### Question 3

The power of (2,1) with respect to the circle 2x2+2y2−8x−6y+k = 0 is positive if

0<k<12

-12<k<12

k>12

k<12

#### SOLUTION

Solution :C

2x2+2y2−8x−6y+k = 0

⇒x2+y2−4x−3y+(k2) = 0

4+1−8−3+k2>0

k2−6>0→k>12

### Question 4

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

(x−2)2+(y−3)2=12

(x−2)2+(y−3)2=4

(x−2)2+(y−3)2=10

(x−2)2+(y−3)2=8

#### SOLUTION

Solution :D

Perpendicular distance from centre to tangent = radius

r=|2+3−9|√2=4√2=4√22=2√2 Equation of circle is (x−2)2+(y−3)2=8

### Question 5

If the line x + 2by + 7 = 0 is diameter of the circle x2+y2−6x+2y=0, then b =

#### 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

### Question 6

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 α−β=2γ, where γ is a constant. The locus of the point of intersection of AP and BQ is

x2−y2−2aytanγ=a2

x2+y2−2aytanγ=a2

x2+y2+2aytanγ=a2

x2−y2+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(cosβ−1)(x−a)⇒y=−cot(β2)(x−a)→(2)

From (1) and (2), tan(α2)=yα+x,tan(β2)=a−xy

Now α−β=2γ

⇒tanγ=tan(α/2)−tan(β/2)(a+x)y+(a−x)y=yα+x−a−xy1+yα+x.a−xy⇒tanγ=tan(α/2)−tan(β/2)1+tan(α/2)tan(β/2)=x2+y2−a22ay⇒x2+y2−2aytanγ=a2 which is required locus.

### Question 7

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

#### 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 (0,1√3).

Thus the equation of the circumcenter is,

(x−0)2+(y−1√3)2=(1−0)2+(0−1√3)2

⇒x2+y2−(2√3)y+13=1+13

⇒3x2+3y2−2√3y=3

### Question 8

The angle at which the circles (x−1)2+y2= 10 and x2+(y−2)2=5 intersect is

π6

π4

π3

π2

#### SOLUTION

Solution :B

r1=√10r2=√5d=√(1)2+(2)2=√5cosθ=r21+r22−d22r1r2=10+5−52√10√5=102×5×√2=1√2

### Question 9

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

#### 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+(K–1)2=K2

h2+2h+2–2K=0

If h is real the Δ≥0

4- 4.1. (2 – 2K) ≥0

= 4 - 8 + 8k ≥0

= 8k ≥4

= k≥12

### Question 10

The number of common tangents to the circles x2+y2+2x+8y−23=0 and x2+y2−4x−10y+19=0 is

1

2

3

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.

### Question 11

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 =

#### 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+y21−a2 = 6, √x22+y22−a2= 7

⇒ x21+y21−a2 = 36

x22+y22−a2 = 49

⇒ x21+y21+x21+y22−2a2 = 85

⇒ x21+y21+x21+y22−2

(x1 x2 + y1 y2) = 85 [From (1)]

⇒ (x21+x22−2x1x2) + (y21+y22−2y1y2) = 85 ⇒ (x1−x2)2 + (y1−y2)2 = 85 ⇒ AB2 = 85 ⇒ AB = √85

### Question 12

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

#### 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.

⇒ √(a−1)2+(b+2)2 = |4a + 3b − 23|√42 + 32

⇒ √(a−1)2+(2a−14+2)2 = |4a+3(2a−14)−23|5 [From (1)]

⇒ 5√a2−2a+1+4a2−48a+144 = |10a−65| ⇒ 5√5a2−50a+145 = 5|2a−13|

⇒ 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 = √(4−1)2+(−6+2)2 = √9+16 = 5

The circle equation is (x−4)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 = √(−6−1)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

### Question 13

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

#### SOLUTION

Solution :C

x2+y2–(y–mx)2=0

Coefficient x2 +Coefficient y2=0

m=+1 or -1

### Question 14

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

### Question 15

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

#### SOLUTION

Solution :B

Radius = Distance from the centre C(1, 2) to x + y + k = 0

⇒ √1+4−3 = |1 + 2 + k|√2 ⇒ |k+3| = 2 ⇒ k + 3 = ± 2 ⇒ k = -1, - 5