Let $C^{\prime}$ be the point on minor arc $BCD$ such that $BC^{\prime}=12$ and $C^{\prime}D=5$, and write $AC^{\prime}=BD=C^{\prime}E=x, AD=y$, and $BD=z$. Ptolemy applied to quadrilaterals $ABC^{\prime}D, BC^{\prime}DE$, and $ABDE$ gives $$\begin{aligned} & x^{2}=12y+5^{2} \\ & x^{2}=5z+12^{2} \\ & yz=14x+5 \cdot 12 \end{aligned}$$ Then $$\left(x^{2}-5^{2}\right)\left(x^{2}-12^{2}\right)=5 \cdot 12yz=5 \cdot 12 \cdot 14x+5^{2} \cdot 12^{2}$$ from which $x^{3}-169x-5 \cdot 12 \cdot 14=0$. Noting that $x>13$, the rational root theorem leads quickly to the root $x=15$. Then triangle $BCD$ has area $\sqrt{16 \cdot 1 \cdot 4 \cdot 11}=8\sqrt{11}$ and circumradius $R=\frac{5 \cdot 12 \cdot 15}{4 \cdot 8\sqrt{11}}=$ $\frac{225\sqrt{11}}{88}$.