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^2

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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$$10x - 10y + c + 12 = 0$ $(i)$If a circle bisects the circumference of another circle,the common chord must pass through the center of the circle being bisected.The center of the second circle $x^2+y^2-6x+4y-12=0$ is $(3, -2)$.Substituting $(3, -2)$ into equation $(i)$:$10(3) - 10(-2) + c + 12 = 0$$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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