SATHEE: Circle 3 Question 11

Circle 3 Question 11

11. The circle $C_{1} : x^{2} + y^{2} = 3$ with centre at $O$ intersects the parabola $x^{2} = 2 y$ at the point $P$ in the first quadrant. Let the tangent to the circle $C_{1}$ at $P$ touches other two circles $C_{2}$ and $C_{3}$ at $R_{2}$ and $R_{3}$ , respectively. Suppose $C_{2}$ and $C_{3}$ have equal radii $2 \sqrt{3}$ and centres $Q_{2}$ and $Q_{3}$ , respectively. If $Q_{2}$ and $Q_{3}$ lie on the $Y$ -axis, then

(a) $Q_{2} Q_{3} = 12$

(2016 Adv.)

(b) $R_{2} R_{3} = 4 \sqrt{6}$

(d) area of the $△ P Q_{2} Q_{3}$ is $4 \sqrt{2}$

Assertion and Reason

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Answer:

Correct Answer: 11. $(a, b, c)$

Solution:

Given, $C_{1} : x^{2} + y^{2} = 3$ intersects the parabola $x^{2} = 2 y$ .

On solving $x^{2} + y^{2} = 3$ and $x^{2} = 2 y$ , we get

$\begin{array}{lc} y^{2} + 2 y = 3 \\ \Rightarrow & y^{2} + 2 y - 3 = 0 \\ \Rightarrow & (y + 3) (y - 1) = 0 \\ ∴ & y = 1, - 3 [neglecting y = - 3, as - \sqrt{3} \leq y \leq \sqrt{3}] \\ ∴ & y = 1 \Rightarrow x = \pm \sqrt{2} \\ \Rightarrow & P (\sqrt{2}, 1) \in I quadrant \end{array}$

Equation of tangent at $P (\sqrt{2}, 1)$ to $C_{1} : x^{2} + y^{2} = 3$ is

$\sqrt{2} x + 1 \cdot y = 3$

Now, let the centres of $C_{2}$ and $C_{3}$ be $Q_{2}$ and $Q_{3}$ , and tangent at $P$ touches $C_{2}$ and $C_{3}$ at $R_{2}$ and $R_{3}$ shown as below

Let $Q_{2}$ be $(0, k)$ and radius is $2 \sqrt{3}$ .

$\begin{array}{lc} ∴ & \frac{| 0 + k - 3 |}{\sqrt{2 + 1}} = 2 \sqrt{3} \\ \Rightarrow & | k - 3 | = 6 \\ \Rightarrow & k = 9, - 3 \\ ∴ & Q_{2} (0, 9) and Q_{3} (0, - 3) \\ Hence, & Q_{2} Q_{3} = 12 \end{array}$

$∴$ Option (a) is correct.

Also, $R_{2} R_{3}$ is common internal tangent to $C_{2}$ and $C_{3}$ , and

$∴$

$\begin{aligned} r_{2} & = r_{3} = 2 \sqrt{3} \\ R_{2} R_{3} & = \sqrt{d^{2} - {(r_{1} + r_{2})}^{2}} = \sqrt{12^{2} - (4 \sqrt{3})^{2}} \\ = \sqrt{144 - 48} = \sqrt{96} = 4 \sqrt{6} \end{aligned}$

$∴$ Option (b) is correct.

$∵$ Length of perpendicular from $O (0, 0)$ to $R_{2} R_{3}$ is equal to radius of $C_{1} = \sqrt{3}$ .

$∴$ Area of $△ O R_{2} R_{3} = \frac{1}{2} \times R_{2} R_{3} \times \sqrt{3} = \frac{1}{2} \times 4 \sqrt{6} \times \sqrt{3} = 6 \sqrt{2}$

$∴$ Option (c) is correct.

Also, area of $△ P Q_{2} Q_{3} = \frac{1}{2} Q_{2} Q_{3} \times \sqrt{2} = \frac{\sqrt{2}}{2} \times 12 = 6 \sqrt{2}$

$∴$ Option (d) is incorrect.

Circle 3 Question 11

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Circle 3 Question 11

Assertion and Reason

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Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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