Circles(11th And 12th > Mathematics ) Questions and answers for exam Preparation

11th And 12th > Mathematics

CIRCLES MCQs

Total Questions : 60 | Page 1 of 6 pages

Question 1.

If $(m_{i}, \frac{1}{m_{i}})$ , i = 1, 2, 3, 4 are con - cyclic points, then the value of $m_{1} m_{2} m_{3} m_{4}$ is

1
-1
0
2

Discuss Question

Answer: Option A. -> 1
:
A

Let equation of circle be $x^{2} + y^{2}$ + 2gx + 2fy + c = 0. If $(m, \frac{1}{m})$ lies on this circle, then

$m^{2} + \frac{1}{m^{2}} + 2 g m + 2 f \frac{1}{m} + c = 0$

or $m^{4} + 2 g m^{3} + 2 f m + c m^{2} + 1 = 0$

This is a fourth degree equation in m having $m_{1}, m_{2}, m_{3}, m_{4}$ as its roots.

Therefore, $m_{1} m_{2} m_{3} m_{4}$ = product of roots = $\frac{1}{1}$ = 1.

Question 2.

The equation of the circle whose centre is (1, -3) and which touches the line 2x - y - 4 = 0 is

$5 x^{2} + 5 y^{2} - 10 x + 30 y + 49 = 0$
$5 x^{2} + 5 y^{2} + 10 x - 30 y + 49 = 0$
$5 x^{2} + 5 y^{2} - 10 x + 30 y - 49 = 0$
None of these

Discuss Question

Answer: Option A. ->

5 x^{2} + 5 y^{2} - 10 x + 30 y + 49 = 0

:
A
Radius of circle the required circle =

∣ ∣ ∣ \frac{2 + 3 - 4}{\sqrt{5}} ∣ ∣ ∣ = \frac{1}{\sqrt{5}}

Therefore the equation is,

(x - 1)^{2} + (y + 3)^{2} = \frac{1}{5}

i.e,

x^{2} + y^{2} - 2 x + 6 y + 1 + 9 = \frac{1}{5}

i.e,

5 x^{2} + 5 y^{2} - 10 x + 30 y + 49 = 0.

Question 3.

The centres of the circles $x^{2} + y^{2} = 1,$ $x^{2} + y^{2} + 6 x - 2 y = 1$ and $x^{2} + y^{2} - 12 x + 4 y = 1$ are

same
Collinear
Non–collinear
None of these

Discuss Question

Answer: Option B. -> Collinear
:
B
Centre are (0, 0), (-3, 1) and (6, -2) and a line passing through any two points say (0, 0) and (-3, 1) is

y = - \frac{1}{3} x

and point (6, -2) lies on it. Hence points are collinear.

Question 4.

The locus of the centre of a circle which touch the circles $x^{2} + y^{2} - 6 x - 6 y + 14 = 0$ externally and also the Y - axis is given by

$x^{2} - 6 y - 7 y + 14 = 0$
$x^{2} - 10 x - 6 y + 14 = 0$
$y^{2} - 6 x - 10 y + 14 = 0$
$y^{2} - 10 x - 6 y + 14 = 0$

Discuss Question

Answer: Option D. ->

y^{2} - 10 x - 6 y + 14 = 0

:
D

Let (x, y) be centre of circle which touch y - axis and given circle.
At any point, the radius of this circle will be equal to 'x' units. Since this circle is touching the given circle which has a radius of two units,
$∴$ Distance between centres of the two circles = 2 + x

$\Rightarrow \sqrt{(x - 3)^{2} + (y - 3)^{2}} = 2 + x$

$y^{2} - 10 x - 6 y + 14 = 0$

Question 5.

The area of the circle whose centre is at (1, 2) and which passes through the point (4, 6) is

$5 π$
$10 π$
$25 π$
None of these

Discuss Question

Answer: Option C. ->

25 π

:
C
Radius =

\sqrt{(1 - 4)^{2} + (2 - 6)^{2}} = 5

Hence the area is given by

π r^{2} = 25 π

sq. units.

Question 6.

If a circle whose centre is (1, –3) touches the line 3x – 4y –5 = 0, then the radius of the circle is

2
4
$\frac{5}{2}$
$\frac{7}{2}$

Discuss Question

Answer: Option A. -> 2
:
A
Radius = perpendicular distance from (1, -3) to the given line 3x - 4y - 5 = 0.
i.e.

∣ ∣ ∣ \frac{3 + 12 - 5}{\sqrt{5^{2}}} ∣ ∣ ∣ = 2.

Question 7.

If a straight line through $C (- \sqrt{8}, \sqrt{8})$ making an angle $135^{\circ}$ with the x-axis cuts the circle $x = 5 c o s θ, y = 5 s i n θ$ in points A and B, then length of segment AB is

5
10
15
$15 \sqrt{2}$

Discuss Question

Answer: Option B. -> 10
:
B
Inclination of the line

\leftarrow \to A B

135^{\circ} \Rightarrow S l o p e = t a n 135^{\circ} = - 1

Equation of

\leftarrow \to A B

y - \sqrt{8} = - 1 (x + \sqrt{8}) \Rightarrow x + y = 0

x + y = 0 passes through the centre of the circle

x^{2} + y^{2} = 25

∴

Length of the chord AB = Diameter of the circle =

2 \times 5 = 10

Question 8.

If $P (2, 8)$ is an interior point of a circle $x^{2} + y^{2} - 2 x + 4 y - p = 0$ which neither touches nor intersects the axes, then set for $p$ is -

$p < - 1$
$p < - 4$
$p > 96$
$p \in ϕ$

Discuss Question

Answer: Option D. ->

p \in ϕ

:
D
Since the point P is interior to the circle,

S_{1}

< 0

= (2^{2}) + (8^{2}) - 2. (2) + 4 (8) - p < 0

= 96 - p < 0

= p > 96

Also given that the circle doesn't touches any of the axes.
So,

g^{2} - c

< 0

f^{2} - c

< 0

g^{2} - c

< 0

= 1 + p < 0

= p < - 1

Also,

f^{2} - c

< 0

= 4 + p < 0

= p < - 4

Since

p < - 4

and

p > 96

is not possible, the set for p will be a null set meaning it'll not have any element in it.

Question 9.

Tangents are drawn from any point on the circle $x^{2} + y^{2} = R^{2}$ to the circle $x^{2} + y^{2} = r^{2}$ . If the line joining the points of intersection of these tangents with the first circle also touch the second, then R equals

$\sqrt{2} r$
$2 r$
$\frac{2 r}{2 - \sqrt{3}}$
$\frac{4 r}{3 - \sqrt{5}}$

Discuss Question

Answer: Option B. ->

2 r

:
B

Tangents Are Drawn From Any Point On The Circle X2+y2=R2 To ...

Δ A B C

is equilateral
Circum radius = 2 (in radius)
R = 2r

Question 10.

The two circles $x^{2} + y^{2} + 2 a x + c = 0 a n d x^{2} + y^{2} + 2 b y + c = 0$ touch if $\frac{1}{a^{2}} + \frac{1}{b^{2}} =$

$\frac{1}{c^{2}}$
$c$
$\frac{1}{c}$
$c^{2}$

Discuss Question

Answer: Option C. ->

\frac{1}{c}

:
C

We can write the equation of common tangent, with the help of radical axis.
Since two circles are touching each other the common tangent will be the radical axis, and the equation of the radical axis we can write -
S = S'
$x^{2} + y^{2} + 2 a x + c = x^{2} + y^{2} + 2 b y + c = 0$
So, Equation of Common tangent is ax – by = 0.

Now, we can find the perpendicular distance from the center of the first circle to the tangent which also should be equal to radius of that circle.
Perpendicular distance from (-a,0) to the tangent = $\frac{a^{2}}{\sqrt{a^{2} + b^{2}}}$
Radius of the same circle will be = $\sqrt{a^{2} - c}$
Therefore $\frac{a^{2}}{\sqrt{a^{2} + b^{2}}} = \sqrt{a^{2} - c}$