If  $C_1:x^2+y^2-20x+64=0$ and $C_2:x^2+y^2+30x+144=0$ , then the length of the shortest line segment PQ which touches $C_1$  at P and $C_2$ at Q is

The centers are (10, 0) and (-15, 0) and the radii are $r_1=6$  and  $r_2=9$.
Also,  $d=25,r_1+r_2<d$.

So, the circles are neither intersecting nor touching. Therefore,
$\begin{array}{l}PQ=\sqrt{d^2-{(r_1+r_2)}^2}\\ =\sqrt{625-225}\\ =20\end{array}$