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Question

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


A
16
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B
24
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C
-42
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D
-62
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Solution

The correct option is D -62
$$S_1=x^2 \, + \, y^2 \, + \, 4x \, - \, 6y \, + \, c \, = \, 0$$ bisects the circumference of the circle $$S_2=x^2 \, + \, y^2 \, - \, 6x \, + \, 4y \, - \, 12 \, = \, 0.$$
Common chord of both the circles is $$S_1-S_2.$$
$$S_1-S_2=10x-10y+(c+12)=0$$
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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