If the circle $x^{2} + y^{2} + 4 x - 6 y + c = 0$ bisects the circumference of the circle $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ , then c =

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-42

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-62

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Solution

The correct option is D -62
$S_{1} = x^{2} + y^{2} + 4 x - 6 y + c = 0$ bisects the circumference of the circle $S_{2} = x^{2} + y^{2} - 6 x + 4 y - 12 = 0.$
Common chord of both the circles is $S_{1} - S_{2} .$
$S_{1} - S_{2} = 10 x - 10 y + (c + 12) = 0$
This chord will pass through point $(3, - 2)$ i.e. centre of $S_{2}$ .
So, $10 (3) - 10 (- 2) + c + 12 = 0$
$62 + c = 0$
$c = - 62$

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