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
B
24
C
-42
D
-62

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, $$10(3) - 10(-2) + c+ 12 = 0$$$$62+ c = 0$$$$c=-62$$Maths

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