Let the line pass through the point of the intersection P (in the first quadrant) of the circle and the parabola . Let the line L touch two circles and of equal radius . If the centres and of the circle and lie on the y-axis, then the square of th

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Solution

Q.
Let the line $L : \sqrt{2} x + y = α$ pass through the point of the intersection P (in the first quadrant) of the circle $x^{2} + y^{2} = 3$ and the parabola $x^{2} = 2 y$ . Let the line L touch two circles $C_{1}$ and $C_{2}$ of equal radius $2 \sqrt{3}$ . If the centres $Q_{1}$ and $Q_{2}$ of the circle $C_{1}$ and $C_{2}$ lie on the y-axis, then the square of the areas of the triangle $P Q_{1} Q_{2}$ is equal to __________. [2024]

Ans.
(72)

We have, $L : \sqrt{2} x + y = α$

$x^{2} + y^{2} = 3$ ... (i)

$x^{2} = 2 y$ ... (ii)

Equation (i) and (ii) intersect at $(\sqrt{2}, 1)$ in first quadrant.

$∴ P = (\sqrt{2}, 1) and α = \sqrt{2} \times \sqrt{2} + 1 = 3$

$\Rightarrow α = 3$

Radius of Circle $C_{1}$ and $C_{2} = 2 \sqrt{3}$

We centre $C_{1}$ & $C_{2}$ of two circle are $(0, y_{1})$ and $(0, y_{2})$ respectively.

We know that length of perpendicular from centre to the tangent = Radius of circle

$∴$ $| \frac{\sqrt{2} \times 0 + y - 3}{\sqrt{3}} |$ $= 2 \sqrt{3}$

$\Rightarrow | y - 3 | = 6 \Rightarrow y = 9, - 3$

$∴ y_{1} = - 3, y_{2} = 9$

$∴$ Centre of $C_{1}$ and $C_{2}$ are $Q_{1} (0, - 3)$ and $Q_{2} (0, 9)$ respectively.

$Q_{1} Q_{2} = 12$ units, Height of triangle $= \sqrt{2}$ units

A = Area of triangle $P Q_{1} Q_{2} = \frac{1}{2} \times 12 \times \sqrt{2}$

$[∵ Area of △ P Q_{1} Q_{2} = \frac{1}{2} \times Q_{1} Q_{2} \times height]$

$= 6 \sqrt{2} sq . units$

$\Rightarrow A^{2} = 72$ .

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