smc

Let $r$ denote the radius of circle $C_1$. Note that quadrilateral $ZYOX$ is cyclic. By Ptolemy's Theorem, we have $11XY=13r+7r$ and $XY=20r/11$. Let $t$ be the measure of angle $YOX$. Since $YO=OX=r$, the law of cosines on triangle $YOX$ gives us $\cos t=-79/121$. Again since $ZYOX$ is cyclic, the measure of angle $YZX=180-t$. We apply the law of cosines to triangle $ZYX$ so that $X{Y}^2=7^2+13^2-2(7)(13)\cos (180-t)$. Since $\cos (180-t)=-\cos t=79/121$ we obtain $X{Y}^2=12000/121$. But$X{Y}^2=400r^2/121$ so that $r=\boxed{\sqrt{30}}$.