Given that circle $x^2+y^2+8 x-4 y+c=0$ touch the circle $x^2+y^2+2 x+4 y-11=0$


where
$\begin{gathered}
C_1=(-4,2) \\
r_1=\sqrt{16+4-c}=\sqrt{20-c} \\
C_2=(-1,-2)
\end{gathered}$
and $r_2=\sqrt{1+4+11}=4$
$\therefore$ From Eq. (i),
$\begin{aligned}
& \sqrt{(-4+1)^2+(2+2)^2} \\
& =\sqrt{20-c}+4 \\
& \Rightarrow \quad 5=\sqrt{20-c}+4 \\
& \Rightarrow \quad c=19 \\
&
\end{aligned}$
Also, the circle $x^2+y^2+8 x-4 y+c=0$ cuts the circles $x^2+y^2-6 x+8 y+k=0$ orthogonally, then $c_1 \rightarrow(-4,2)$
$\begin{aligned} & C_3 \rightarrow(3,-4) \\ & \left(C_1 C_3\right)^2=\left(r_1\right)^2+\left(r_3\right)^2 \\ & \text { where, } \\ & r_1=\sqrt{16+4-c} \\ & r_3=\sqrt{9+16-k} \\ & \Rightarrow \\ & \left\{(-4-3)^2+(2+4)^2\right\}=(20-c)+(25-k) \\ & \Rightarrow \quad 49+36=45-k-c \\ & \Rightarrow \quad k+c=-40 \\ & \Rightarrow \quad k+19=-40 \\ & \Rightarrow \quad k=-59 \\ & \end{aligned}$