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 $x^{2} + y^{2} = 9$ can be

$(\frac{1}{2}, \frac{1}{2})$

$(\frac{1}{2}, - \sqrt{2})$

$(\frac{3}{2}, \frac{1}{2})$

$(\frac{1}{2}, \frac{3}{2})$

SOLUTION

Solution : B

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

$x = \frac{0 + 1}{2}$

$= \frac{1}{2}$

$r = \frac{3}{2}$

$\sqrt{x^{2} + y^{2}} = \frac{3}{2}$

$y = \pm \sqrt{2}$

So there are two circles possible.

Question 2

The length of common chord of the circles $(x - a)^{2} + y^{2}$ = $a^{2}$ and $x^{2} + (y - b)^{2}$ = $b^{2}$ is

$2 \sqrt{a^{2} + b^{2}}$

$\frac{a b}{\sqrt{a^{2} + b^{2}}}$

$\frac{2 a b}{\sqrt{a^{2} + b^{2}}}$

$\frac{2 a b}{\sqrt{a^{2} - b^{2}}}$

SOLUTION

Solution : C

Equation of common chord is $a x - b y$ = 0

Now length of common chord

= $2 \sqrt{r_{1}^{2} - p_{1}^{2}}$ = $2 \sqrt{r_{2}^{2} - p_{2}^{2}}$

Where $r_{1}$ and $r_{2}$ are radii of given circles and $p_{1}, p_{2}$ are the perpendicular distances from centres of circles to common chords.

Hence required length = $2 \sqrt{a^{2} - \frac{a^{4}}{a^{2} + b^{2}}}$ = $\frac{2 a b}{\sqrt{a^{2} + b^{2}}}$ .

Question 3

The power of (2,1) with respect to the circle $2 x^{2} + 2 y^{2} - 8 x - 6 y + k$ = 0 is positive if

0<k<12

-12<k<12

k>12

k<12

SOLUTION

Solution : C

$2 x^{2} + 2 y^{2} - 8 x - 6 y + k$ = 0

$\Rightarrow x^{2} + y^{2} - 4 x - 3 y + (\frac{k}{2})$ = 0

$4 + 1 - 8 - 3 + \frac{k}{2} > 0$

$\frac{k}{2} - 6 > 0 \to k > 12$

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

$r = \frac{| 2 + 3 - 9 |}{\sqrt{2}} = \frac{4}{\sqrt{2}} = \frac{4 \sqrt{2}}{2} = 2 \sqrt{2}$ 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 $x^{2} + y^{2} - 6 x + 2 y = 0$ , 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 $\Rightarrow$ 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 $α - β = 2 γ$ , where $γ$ is a constant. The locus of the point of intersection of AP and BQ is

$x^{2} - y^{2} - 2 a y t a n γ = a^{2}$

$x^{2} + y^{2} - 2 a y t a n γ = a^{2}$

$x^{2} + y^{2} + 2 a y t a n γ = a^{2}$

$x^{2} - y^{2} + 2 a y t a n γ = a^{2}$

SOLUTION

Solution : B

Coordinates of A are (–a, 0) and of P are $(a c o s α, a s i n α)$
$∴$ Equation of AP is $y = \frac{a s i n α}{a (c o s α + 1)} (x + a) \Rightarrow y = t a n (\frac{α}{2}) (x + a) \to (1)$
Similarly equation of BQ = $y = \frac{a s i n β}{a (c o s β - 1)} (x - a) \Rightarrow y = - c o t (\frac{β}{2}) (x - a) \to (2)$
From (1) and (2), $t a n (\frac{α}{2}) = \frac{y}{α + x}, t a n (\frac{β}{2}) = \frac{a - x}{y}$
Now $α - β = 2 γ$
$\Rightarrow t a n γ = \frac{t a n (α / 2) - t a n (β / 2)}{(a + x) y + (a - x) y} = \frac{\frac{y}{α + x} - \frac{a - x}{y}}{1 + \frac{y}{α + x} . \frac{a - x}{y}} \Rightarrow t a n γ = \frac{t a n (α / 2) - t a n (β / 2)}{1 + t a n (α / 2) t a n (β / 2)} = \frac{x^{2} + y^{2} - a^{2}}{2 a y} \Rightarrow x^{2} + y^{2} - 2 a y t a n γ = a^{2}$ 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

3 x^{2} + 3 y^{2} - 2 \sqrt{3} y = 3

2 x^{2} + 2 y^{2} - 3 \sqrt{2} y = 3

x^{2} + y^{2} - 2 y = 1

N o n e

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 $A B = A C$
$\Rightarrow \sqrt{1 + b^{2}} = 2$
$\Rightarrow b^{2} = 3$
$\Rightarrow b = \sqrt{3}$ [ $∵$ b>0]
Since the triangle is equilateral, the centre of the circumcircle is at the centroid of the triangle which is $(0, \frac{1}{\sqrt{3}})$ .
Thus the equation of the circumcenter is,
$(x - 0)^{2} + {(y - \frac{1}{\sqrt{3}})}^{2} = (1 - 0) 2 + {(0 - \frac{1}{\sqrt{3}})}^{2}$
$\Rightarrow x^{2} + y^{2} - (\frac{2}{\sqrt{3}}) y + \frac{1}{3} = 1 + \frac{1}{3}$
$\Rightarrow 3 x^{2} + 3 y^{2} - 2 \sqrt{3} y = 3$

Question 8

The angle at which the circles $(x - 1)^{2} + y^{2} = 10 a n d x^{2} + (y - 2)^{2} = 5$ intersect is

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{2}$

SOLUTION

Solution : B

$r_{1} = \sqrt{10} r_{2} = \sqrt{5} d = \sqrt{(1)^{2} + (2)^{2}} = \sqrt{5} c o s θ = \frac{r_{1}^{2} + r_{2}^{2} - d^{2}}{2 r_{1} r_{2}} = \frac{10 + 5 - 5}{2 \sqrt{10} \sqrt{5}} = \frac{10}{2 \times 5 \times \sqrt{2}} = \frac{1}{\sqrt{2}}$

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

- \frac{1}{2} \leq k \leq \frac{1}{2}

k \leq \frac{1}{2}

C. < k <

\frac{1}{2}

k \geq \frac{1}{2}

SOLUTION

Solution : D

Since, circles are passing through the point (–1, 1) and touching X-axis, the equationof circles can be written as -
$(h + 1)^{2} + (K - 1)^{2} = K^{2}$
$h^{2} + 2 h + 2 - 2 K = 0$
If h is real the $Δ \geq 0$
4- 4.1. (2 – 2K) $\geq 0$
= 4 - 8 + 8k $\geq 0$
= 8k $\geq 4$
= $k \geq \frac{1}{2}$

Question 10

The number of common tangents to the circles $x^{2} + y^{2} + 2 x + 8 y - 23 = 0 a n d x^{2} + y^{2} - 4 x - 10 y + 19 = 0$ is

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 - $\sqrt{(3^{2} + 9^{2})}$
= 3 $\sqrt{(} 10)$
Now, the sum their radii will be -
Radius of the first circle is - 2 $\sqrt{(} 10)$
Radius of the second circle is - $\sqrt{(} 10)$
Sum of them will be - 3 $\sqrt{(} 10)$
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 =

A. 5

\sqrt{85}

\frac{\sqrt{85}}{2}

D. none

SOLUTION

Solution : B

Let the equation of the circle be $x^{2}$ + $y^{2} = a^{2}$ and A $(x_{1}, y_{1})$ , $B (x_{2}, y_{2})$
A, B are conjugate points $\Rightarrow x_{1} x_{2} + y_{1} y_{2} = a^{2}$ $\to$ (1)
Length of the tangents from A, B are 6, 7 respectively $\Rightarrow$ $\sqrt{x_{1}^{2} + y_{1}^{2} - a^{2}}$ = 6, $\sqrt{x_{2}^{2} + y_{2}^{2} - a^{2}}$ = 7
$\Rightarrow$ $x_{1}^{2} + y_{1}^{2} - a^{2}$ = 36
$x_{2}^{2} + y_{2}^{2} - a^{2}$ = 49
$\Rightarrow$ $x_{1}^{2} + y_{1}^{2} + x_{1}^{2} + y_{2}^{2} - {2 a}^{2}$ = 85
$\Rightarrow$ $x_{1}^{2} + y_{1}^{2} + x_{1}^{2} + y_{2}^{2} - 2$
(x $_{1}$ x $_{2}$ + y $_{1}$ y $_{2}$ ) = 85 [From (1)]
$\Rightarrow$ $(x_{1}^{2} + x_{2}^{2} - 2 x_{1} x_{2})$ + $(y_{1}^{2} + y_{2}^{2} - 2 y_{1} y_{2})$ = 85 $\Rightarrow$ $(x_{1} - x_{2})^{2}$ + $(y_{1} - y_{2})^{2}$ = 85 $\Rightarrow$ $A B^{2}$ = 85 $\Rightarrow$ AB = $\sqrt{85}$

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

x^{2}

y^{2}

+ 8x + 12y + 27 = 0

x^{2}

y^{2}

- 12y + 27 = 0

x^{2}

y^{2}

- 8x + 12y - 27 = 0

x^{2}

y^{2}

- 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 $\Rightarrow$ 2a - b - 14 = 0 $\Rightarrow$ b = 2a - 14 $\to$ (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.
$\Rightarrow$ $\sqrt{{(a - 1)}^{2} + {(b + 2)}^{2}}$ = $\frac{| 4 a + 3 b - 23 |}{\sqrt{4^{2} + 3^{2}}}$
$\Rightarrow$ $\sqrt{{(a - 1)}^{2} + {(2 a - 14 + 2)}^{2}}$ = $\frac{| 4 a + 3 (2 a - 14) - 23 |}{5}$ [From (1)]
$\Rightarrow$ 5 $\sqrt{a^{2} - 2 a + 1 + {4 a}^{2} - 48 a + 144}$ = $| 10 a - 65 |$ $\Rightarrow$ 5 $\sqrt{{5 a}^{2} - 50 a + 145}$ = 5 $| 2 a - 13 |$
$\Rightarrow$ 5a $^{2}$ - 50a + 145 = (2a - 13) $^{2}$ = 4a $^{2}$ - 52a + 169
$\Rightarrow$ a $^{2}$ + 2a - 24 = 0 $\Rightarrow$ (a + 6)(a - 4) $\Rightarrow$ a = 4
Case 1 : a = 4 $\Rightarrow$ b = 2(4) - 14 = - 6. $∴$ Centre C = (4, - 6)
Radius = CP = $\sqrt{{(4 - 1)}^{2} + {(- 6 + 2)}^{2}}$ = $\sqrt{9 + 16}$ = 5
The circle equation is $(x - 4)^{2} + (y + 6)^{2} = 5^{2}$ $\Rightarrow$ $x^{2}$ + $y^{2}$ - 8x + 12y + 27 = 0
Case 2 : a = - 6 $\Rightarrow$ b = 2(- 6) - 14 = 26 $∴$ Centre C = (-6, - 26)
Radius = CP = $\sqrt{{(- 6 - 1)}^{2} + {(- 26 + 2)}^{2}}$ = $\sqrt{49 + 576}$ = $\sqrt{625}$ = 25
The circle equation is $(x + 6)^{2} + (y + 26)^{2} = 25^{2}$ $\Rightarrow$ $x^{2}$ + $y^{2}$ + 12x + 52y + 87 = 0
$∴$ The circles are $x^{2}$ + $y^{2}$ - 8x + 12y + 27 = 0, $x^{2}$ + $y^{2}$ + 12x + 52y + 87 = 0

Question 13

If the chord y = mx + 1 of the circle $x^{2} + y^{2} = 1$ subtends an angle of measure $45^{\circ}$ 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

$x^{2} + y^{2} - (y - m x)^{2} = 0$
Coefficient $x^{2}$ +Coefficient $y^{2} = 0$
m=+1 or -1

Question 14

The equation of the image of the circle $x^{2} + y^{2}$ - 6x - 4y + 12 = 0 by the line mirror x + y - 1 = 0

x^{2} + y^{2}

+ 2x + 4y + 4 = 0

x^{2} + y^{2}

- 2x + 4y + 4 = 0

x^{2} + y^{2}

+ 2x + 4y - 4 = 0

x^{2} + y^{2}

+ 2x - 4y - 4 = 0

SOLUTION

Solution : A

For the circle $x^{2} + y^{2}$ - 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 $x^{2} + y^{2}$ - 6x - 4y + 12 = 0 w.r.t x + y - 1 = 0 is,
$x^{2} + y^{2}$ + 2x + 4y + 4 = 0

Question 15

If x + y + k = 0 is a tangent to the circle $x^{2} + y^{2}$ - 2x - 4y + 3 = 0 then k =

\pm

B. –1, – 5

\pm

D. 4

SOLUTION

Solution : B

Radius = Distance from the centre C(1, 2) to x + y + k = 0
$\Rightarrow$ $\sqrt{1 + 4 - 3}$ = $\frac{| 1 + 2 + k |}{\sqrt{2}}$ $\Rightarrow$ $| k + 3 |$ = 2 $\Rightarrow$ k + 3 = $\pm$ 2 $\Rightarrow$ k = -1, - 5

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

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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