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Angle of Intersection of Two Circles

If the circle...

Question

If the circle $x^{2} + y^{2} + 4 x + 22 y + c = 0$ bisects the circumference of the circle $x^{2} + y^{2} - 2 x + 8 y - d = 0 (c, d > 0),$ then the maximum possible value of $c d$ is

25

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125

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425

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625

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Solution

The correct option is D $625$

Given that $S_{1} = x^{2} + y^{2} + 4 x + 22 y + c = 0$
$S_{2} = x^{2} + y^{2} - 2 x + 8 y - d = 0$
$A B$ is a common chord passing through $C_{2} .$
Equation of $A B : S_{1} - S_{2} = 0$
$\Rightarrow (x^{2} + y^{2} + 4 x + 22 y + c) - (x^{2} + y^{2} - 2 x + 8 y - d) = 0$
$\Rightarrow 6 x + 14 y + c + d = 0$
The common chord passes through centre of $S_{2}$ i.e., $C_{2} (1, - 4)$
$\Rightarrow 6 (1) + 14 (- 4) + c + d = 0$
$\Rightarrow c + d = 50$
Since $A . M . \geq G . M .$
$\frac{c + d}{2} \geq \sqrt{c d}$
$\Rightarrow 25 \geq \sqrt{c d}$
$\Rightarrow c d \leq 625$

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