By the Law of Cosines on triangle $ABC$, we have
\begin{align*}
\cos \angle ABC &= \frac{8^2+15^2-12^2}{2\cdot 8 \cdot 15} = \frac{97}{120}\\
\cos \angle ABC &= \cos(\angle ABC + \angle CAB) \\
\cos \angle ABC &= \cos(\angle BAC) \cos(\angle ABC) - \sin(\angle BAC) \sin(\angle ABC) \\
2 \cos^2 \angle ABC - 1 &= (2 \cos(\angle ABC)^2 - 1) \cos(\angle BAC) - 2 \sin(\angle ABC) \sin(\angle BAC) \cos(\angle ABC) \\
153 \cdot \frac{97}{120} - 1 &= \left(\frac{97}{120}\right)^2 \cos(\angle BAC) - 2 \frac{120}{\sqrt{1441}} \cdot \frac{97}{120} \cdot \frac{77}{120}\\
\frac{11681}{80} &= \left(\frac{97}{120}\right)^2 \cos(\angle BAC) - \frac{5099}{96}\\
\frac{11681}{80} + \frac{5099}{96} &= \left(\frac{97}{120}\right)^2 \cos(\angle BAC)\\
\frac{77909}{240} &= \left(\frac{97}{120}\right)^2 \cos(\angle BAC)\\
160^2 &= 120^2 \cos^2(\angle BAC)\\
160^2 &= 120^2 - 120^2 \sin^2(\angle BAC)\\
s &= \sqrt{120^2 - 120^2 \sin^2(\angle BAC)}
\end{align*}Since $\angle BAC$ is acute,
\[\sin \angle BAC = \sqrt{1 - \cos^2 \angle BAC} = \sqrt{1 - \left( \frac{160}{120} \right)^2} = \frac{3}{\sqrt{7}}.\]Thus,
\[s = \sqrt{120^2 - 120^2 \cdot \frac{9}{7}} = \sqrt{120^2 \cdot \frac{16}{7}} = \boxed{\frac{240}{\sqrt{7}}}.\]
Points $A$, $B$, and $C$ are on a circle such that $AB = 8$, $BC = 15$, and $AC = 12$. Find the radius of the circle.
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