The circle $C_{1} : x^{2} + y^{2} = 3$ with centre at O intersects the curve $x^{2} = 2 y$ 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 centers $Q_{2}$ and $Q_{3}$
Respectively. If $Q_{2}$ and $Q_{3}$ lie on the Y-axis. Then

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area of the $Δ O R_{2} R_{3} i s = 6 \sqrt{2}$

area of the $Δ P Q_{2} Q_{3} i s = 4 \sqrt{2}$

$Q_{2} Q_{3} = 12$

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

answer is A, B, C.

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Detailed Solution

Equation of tangent at $P (\sqrt{2}, 1)$ is $\sqrt{2} x + y - 3 = 0$
If centre of $C_{2}$ at $(0, α)$ and radius equal to
$2 \sqrt{3} = | \frac{α - 3}{\sqrt{3}} | \Rightarrow α = - 3, 9$
A. $Q_{2} Q_{3} = 12$
B. $R_{2} R_{3} =$ length of transverse common tangent
   $= \sqrt{{(Q_{2} Q_{3})}^{2} - {(r_{1} + r_{2})}^{2}} = \sqrt{{(12)}^{2} - {(2 \sqrt{3} + 2 \sqrt{3})}^{2}} = 4 \sqrt{6}$
C. Area of $Δ O R_{2} R_{3}$
   $= \frac{1}{2} \times R_{2} R_{3} \times$ perpendicular distance of O from line
   $= \frac{1}{2} \times 4 \sqrt{6} \times \sqrt{3} = 6 \sqrt{2}$
D. Area of $Δ P Q_{2} Q_{3} = \frac{1}{2} \times 12 \times \sqrt{2} = 6 \sqrt{2}$
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