The common chord of the given circle is
$$
\begin{array}{cc}
& S_1-S_2=0 \\
\Rightarrow & \left(x^2+y^2+4 x-6 y+c\right) \\
& -\left(x^2+y^2-6 x+4 y-12\right)=0 \\
\Rightarrow & 10 x-10 y+c+12=0
\end{array}
$$
Since, circle $x^2+y^2+4 x+6 y+c=0$ bisects the circumference of the circle.
$$
x^2+y^2-6 x+4 y-12=0
$$
Therefore, Eq. (i) passes through the centre of second circle i.e., $(3,-2)$
$$
\begin{aligned}
& \therefore & 10(3)-10(-2)+c+12 & =0 \\
\Rightarrow & & 30+20+c+12 & =0 \\
\Rightarrow & & c & =-62
\end{aligned}
$$