If the circle $x^{2} + y^{2} + 4 x + 22 y + a = 0$ bisects the circumference of the circle $x^{2} + y^{2} - 2 x + 8 y - b = 0$ (where $a, b > 0)$ , then find the maximum value of $(a b)$

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Solution

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Given
$S_{1} : x^{2} + y^{2} + 4 x + 22 y + a = 0$
$S_{2} : x^{2} + y^{2} - 2 x + 8 y + b = 0$
common chord : $S_{1} - S_{2} = 0$
$\Rightarrow 6 x + 14 y + a + b = 0$
Since, common chord passes through $(1, - 4)$
$6 \times 1 + 14 (- 4) + a + b = 0$
$\Rightarrow a + b = 50$
let $y = a b = a (50 - a)$
$\frac{d y}{d x} = 50 - 2 a$
$∴ \frac{d b}{d a} = 0$
$∴ a = 25$
$∴ (a b) m a x = 625$

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