smc

First examine the formula $(x-10{)}^2+y^2=36$, for the circle $C_1$. Its center, $D_1$, is located at (10,0) and it has a radius of $\sqrt{36}$ = 6. The next circle, using the same pattern, has its center, $D_2$, at (-15,0) and has a radius of $\sqrt{81}$ = 9. So we can construct this diagram: Line PQ is tangent to both circles, so it forms a right angle with the radii (6 and 9). This, as well as the two vertical angles near O, prove triangles $D_{2}QO$ and $D_{1}PO$ similar by AA, with a scale factor of $6:9$, or $2:3$. Next, we must subdivide the line $D_2{D}_1$ in a 2:3 ratio to get the length of the segments $D_{2}O$ and $D_{1}O$. The total length is $10-(-15)$, or $25$, so applying the ratio, $D_{2}O$ = 15 and $D_{1}O$ = 10. These are the hypotenuses of the triangles. We already know the length of $D_{2}Q$ and $D_{1}P$, 9 and 6 (they're radii). So in order to find $PQ$, we must find the length of the longer legs of the two triangles and add them. $15^2-9^2=(15-9)(15+9)=6\times 24=144$ $\sqrt{144}=12$ $10^2-6^2=(10-6)(10+6)=4\times 16=64$ $\sqrt{64}=8$ Finally, the length of PQ is $12+8=\boxed{20}$, or (C).