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(C) The given circle is $x^2+y^2+6x-4y-12=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=3, f=-2, c=-12$. Centre $O = (-g, -f) = (-3, 2)$ and radi

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$2x+3y \pm 5\sqrt{26}=0$

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(C) The given circle is $x^2+y^2+6x-4y-12=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=3, f=-2, c=-12$. Centre $O = (-g, -f) = (-3, 2)$ and radius $r = \sqrt{g^2+f^2-c} = \sqrt{9+4+12} = 5$. The locus of the point of intersection of perpendicular tangents to a circle is its director circle. The director circle $S$ has the same centre as the original circle and radius $R = r\sqrt{2} = 5\sqrt{2}$. Thus,the equation of circle $S$ is $(x+3)^2+(y-2)^2 = (5\sqrt{2})^2 = 50$. The line perpendicular to $6x-4y+k=0$ will have the form $4x+6y+C=0$,which can be written as $2x+3y+C'=0$. The distance from the centre $(-3, 2)$ to the tangent $2x+3y+C'=0$ must equal the radius $R = 5\sqrt{2}$. Using the distance formula: $\frac{|2(-3)+3(2)+C'|}{\sqrt{2^2+3^2}} = 5\sqrt{2}$. $\frac{|-6+6+C'|}{\sqrt{13}} = 5\sqrt{2} \Rightarrow |C'| = 5\sqrt{26}$. Therefore,the equation of the tangent is $2x+3y \pm 5\sqrt{26}=0$.

Let the locus of the point of intersection of the perpendicular tangents drawn to the circle $x^2+y^2+6x-4y-12=0$ be the circle $S$. Then the equation of the tangent drawn to $S$ which is perpendicular to the line $6x-4y+k=0$ is

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