$$
\text { } r_1=\sqrt{k^2+4-1} \Rightarrow \sqrt{k^2+3}
$$




and $r_2=\sqrt{16+36-43}=\sqrt{52-43}=\sqrt{9}=3$
common tangent $=S_1-S_2=0$
So, $x^2+y^2+2 k x-4 y+1-x^2-y^2+8 x+12 y-43=0$
$$
(2 k+8) x+8 y-42=0
$$
Perpendicular drawn from any circle is equal to respective circle
$$
\begin{array}{lc}
\text { So, } \quad \frac{(2 k+8) 4+48-42}{\sqrt{(2 k+8)^2+64}}=3 \\
\Rightarrow \quad \frac{8 k+32+6}{\sqrt{4 k^2+64+32 k+64}}=3 \\
\Rightarrow \quad 8 k+38=3 \sqrt{4 k^2+32 k+128} \\
\Rightarrow \quad[8 K+38]^2=9\left(4 k^2+32 k+128\right] \\
k=-1 \text { satisfy the Eq. (i), so } k=-1 .
\end{array}
$$