If the circle x^2+y^2+4x-6y+c=0 bisects the c | vedclass

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(D) The common chord of the two circles $S_1 = x^2+y^2+4x-6y+c=0$ and $S_2 = x^2+y^2-6x+4y-12=0$ is given by $S_1 - S_2 = 0$. $(x^2+y^2+4x-6y+c) - (x^

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$-62$

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(D) The common chord of the two circles $S_1 = x^2+y^2+4x-6y+c=0$ and $S_2 = x^2+y^2-6x+4y-12=0$ is given by $S_1 - S_2 = 0$. $(x^2+y^2+4x-6y+c) - (x^2+y^2-6x+4y-12) = 0$ $10x - 10y + c + 12 = 0$ Since the first circle bisects the circumference of the second circle,the common chord must pass through the center of the second circle. The center of the second circle $x^2+y^2-6x+4y-12=0$ is $(-g, -f) = (3, -2)$. Substituting $(3, -2)$ into the equation of the common chord: $10(3) - 10(-2) + c + 12 = 0$ $30 + 20 + c + 12 = 0$ $62 + c = 0$ $c = -62$

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 $c$ is equal to

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