If the circle $x^{2} + y^{2} + 6 x + 8 y + a = 0$ bisects the circumference of the circle $x^{2} + y^{2} + 2 x - 6 y - b = 0$ , then $a + b$ is equal to

38

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- 38

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42

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None of these

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Solution

The correct option is A $- 38$
Given circles are
$S_{1} : x^{2} + y^{2} + 6 x + 8 y + a = 0$ ...(1)
and $S_{2} : x^{2} + y^{2} + 2 x - 6 y - b = 0$ ...(2)
The equation of common chord of the two circles is
$S_{1} - S_{2} = 0 \Rightarrow 4 x + 14 y + (a + b) = 0$ ...(3)
Since the circle $S_{1}$ bisects the circumference of the circle $S_{2},$ therefore, (1) passes through the center of the circle $S_{1}$ i.e. $(- 1, 3)$
$∴ 4 (- 1) + 14 (3) + a + b = 0 \Rightarrow a + b = - 38$

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