smc

By the Two Tangent Theorem, the segments drawn from point $P$ tangent to circle $O$ are congruent. Let them have length $p$. Similarly, the two segments drawn from point $Q$ tangent to circle $O^{\prime }$ are congruent. Let them have length $q$. Let the length of the two external common tangents be $x$. We know that the two segments drawn from point $P$ tangent to circle $O^{\prime }$ are congruent. One of these tangents (the one on the external common tangent) has length $x-p$, so the other (the one on the internal common tangent) also has length $x-p$ Doing the same thing for point $Q$ and circle $O$ shows that the segments drawn from $Q$ tangent to circle $O$ has length $x-q$. Thus, $PQ=\frac{x-p+x-q+p+q}{2}=\frac{2x}{2}=x$, because the numerator of this fraction counts the length of $\overline{PQ}$ twice. Because $x$ is the length of the external common tangent, our answer is $\boxed{\text{always equal to the length of an external common tangent}}$.