JEE Advanced · Mathematics · 12. Circle

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 $Q_{2} Q_{3} = 12$
B $R_{2} R_{3} = 4 \sqrt{6}$
C area of the triangle $O R_{2} R_{3}$ is $6 \sqrt{2}$
D area of the triangle $P Q_{2} Q_{3}$ is $4 \sqrt{2}$

Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

On solving

x^{2} + y^{2} = 3

and

x^{2} = 2 y

we get point

P (\sqrt{2}, 1)

Equation of tangent at P

\sqrt{2} x + y = 3

Let

Q_{2}

be (0, k) and radius is

2 \sqrt{3}

∴

|\frac{\sqrt{2} (0) + k - 3}{\sqrt{2 + 1}}| = 2 \sqrt{3}

∴

k = 9, - 3

Q_{3}

Q_{2} (0, 9)

and

Q_{3}

(0, - 3)

{\begin{matrix} C_{2} : {(x - 0)}^{2} + {(y - 9)}^{2} = 12 \\ C_{3} : {(x - 0)}^{2} + {(y + 3)}^{2} = 12 \end{matrix}

Hence

Q_{2} Q_{3} = 12

{\Rightarrow a}^{2} + 12 = 36

\Rightarrow a = 2 \sqrt{6}

R_{2} R_{3} = 2 a = 4 \sqrt{6}

Perpendicular distance of origin O from

R_{2} R_{3}

is equal to distance of O from tangent

\sqrt{2} x + y = 3

which is same as radius of circle

C_{1} = \sqrt{3}

Hence area of

Δ O R_{2} R_{3} = \frac{1}{2} \times (R_{2} R_{3}) \sqrt{3} = \frac{1}{2} . 4 \sqrt{6} . \sqrt{3} = 6 \sqrt{2}

Perpendicular Distance of P from

Q_{2} Q_{3} = \sqrt{2}

∴

Area of

Δ P Q_{2} Q_{3} = \frac{1}{2} \times 12 \times \sqrt{2} = 6 \sqrt{2}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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