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(C) The equation of a circle passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + \lambda S_2 = 0$.$(x^2+y^2+6x+4y-1

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$x^2+y^2-2y-12=0$

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(C) The equation of a circle passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + \lambda S_2 = 0$.$(x^2+y^2+6x+4y-12) + \lambda(x^2+y^2-4x-6y-12) = 0$$(1+\lambda)x^2 + (1+\lambda)y^2 + (6-4\lambda)x + (4-6\lambda)y - 12(1+\lambda) = 0$Dividing by $(1+\lambda)$,we get $x^2 + y^2 + 2\left(\frac{3-2\lambda}{1+\lambda}\right)x + 2\left(\frac{2-3\lambda}{1+\lambda}\right)y - 12 = 0$.The radius $r$ of the circle $x^2+y^2+2gx+2fy+c=0$ is $\sqrt{g^2+f^2-c}$.Given $r = \sqrt{13}$,so $g^2+f^2-c = 13$.$\left(\frac{3-2\lambda}{1+\lambda}\right)^2 + \left(\frac{2-3\lambda}{1+\lambda}\right)^2 - (-12) = 13$$\frac{9+4\lambda^2-12\lambda + 4+9\lambda^2-12\lambda}{(1+\lambda)^2} = 1$$13\lambda^2 - 24\lambda + 13 = 1 + 2\lambda + \lambda^2$$12\lambda^2 - 26\lambda + 12 = 0 \Rightarrow 6\lambda^2 - 13\lambda + 6 = 0$.Solving for $\lambda$: $(2\lambda-3)(3\lambda-2) = 0$,so $\lambda = \frac{3}{2}$ or $\lambda = \frac{2}{3}$.For $\lambda = \frac{2}{3}$,the equation is $x^2+y^2+2x-12=0$.For $\lambda = \frac{3}{2}$,the equation is $x^2+y^2-2y-12=0$.

The equation of the circle passing through the points of intersection of the circles $x^2+y^2+6x+4y-12=0$ and $x^2+y^2-4x-6y-12=0$ and having radius $\sqrt{13}$ is

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