Circle $C_1$ has center $O$ on circle $C_2$; they meet at $X,Y$. Point $Z$ on $C_2$ (exterior to $C_1$) has $XZ=7,OZ=8,YZ=4$. What is the radius of $C_1$?

1. With $\angle OZX=\angle OZY=\alpha$: $x^2=OZ^2+XZ^2-2\cdot OZ\cdot XZ\cos\alpha=OZ^2+YZ^2-2\cdot OZ\cdot YZ\cos\alpha$.
2. This gives $x^2=113-112\cos\alpha$ and $x^2=80-64\cos\alpha$. Multiplying the first by $4$ and the second by $7$ and subtracting eliminates $\cos\alpha$: $3x^2=7(80)-4(113)=108$.
3. $x^2=36$, so $x=\boxed6$. (Check: $\cos\alpha=\frac{80-36}{64}=\frac{11}{16}$, a valid angle.)