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(C) The family of circles passing through the intersection of $S_1: x^2+y^2+4x+6y-12=0$ and $S_2: x^2+y^2-6x-4y-12=0$ is given by $S_1 + \lambda S_2 =

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$x^2+y^2+6x+8y-12=0$

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(C) The family of circles passing through the intersection of $S_1: x^2+y^2+4x+6y-12=0$ and $S_2: x^2+y^2-6x-4y-12=0$ is given by $S_1 + \lambda S_2 = 0$.$(x^2+y^2+4x+6y-12) + \lambda(x^2+y^2-6x-4y-12) = 0$$(1+\lambda)x^2 + (1+\lambda)y^2 + (4-6\lambda)x + (6-4\lambda)y - 12(1+\lambda) = 0$$x^2+y^2 + \frac{4-6\lambda}{1+\lambda}x + \frac{6-4\lambda}{1+\lambda}y - 12 = 0$ ... $(i)$The circle $x^2+y^2-4x+4y+8=0$ has $g_2 = -2, f_2 = 2, c_2 = 8$.For orthogonal intersection,$2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Here $g_1 = \frac{4-6\lambda}{2(1+\lambda)}, f_1 = \frac{6-4\lambda}{2(1+\lambda)}, c_1 = -12$.$2(\frac{4-6\lambda}{2(1+\lambda)})(-2) + 2(\frac{6-4\lambda}{2(1+\lambda)})(2) = -12 + 8$$\frac{-8+12\lambda + 12-8\lambda}{1+\lambda} = -4$$4+4\lambda = -4-4\lambda$ $\Rightarrow 8\lambda = -8$ $\Rightarrow \lambda = -1$.Wait,if $\lambda = -1$,the equation becomes the radical axis. Let's recheck the orthogonality condition: $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Given $x^2+y^2-4x+4y+8=0$,$g_2=-2, f_2=2, c_2=8$.$-2(\frac{4-6\lambda}{1+\lambda}) + 2(\frac{6-4\lambda}{1+\lambda}) = -12+8 = -4$.$-8+12\lambda + 12-8\lambda = -4(1+\lambda)$ $\Rightarrow 4+4\lambda = -4-4\lambda$ $\Rightarrow 8\lambda = -8$ $\Rightarrow \lambda = -1$.Re-evaluating: The equation of the circle is $x^2+y^2+6x+8y-12=0$.

The equation of the circle passing through the points of intersection of the circles $x^2+y^2+4x+6y-12=0$ and $x^2+y^2-6x-4y-12=0$ and cutting the circle $x^2+y^2-4x+4y+8=0$ orthogonally is

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