In triangle $ABC,$ $AB = 3$ and $AC = 5.$  Let $O$ be the circumcenter of triangle $ABC.$  Find $\overrightarrow{OA} \cdot \overrightarrow{BC}.$
Let $\mathbf{a} = \overrightarrow{OA},$ $\mathbf{b} = \overrightarrow{OB},$ and $\mathbf{c} = \overrightarrow{OC}.$  Then
\[\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}.\]Similarly, $\overrightarrow{AC} = \mathbf{c} - \mathbf{a}$ and $\overrightarrow{BC} = \mathbf{c} - \mathbf{b}.$  We then want to compute
\[\overrightarrow{OA} \cdot \overrightarrow{BC} = \mathbf{a} \cdot (\mathbf{c} - \mathbf{b}) = \mathbf{a} \cdot \mathbf{c} - \mathbf{a} \cdot \mathbf{b}.\][asy]
unitsize(2 cm);

pair A, B, C, O;

A = dir(100);
B = dir(200);
C = dir(340);
O = (0,0);

draw(Circle(O,1));
draw(A--B--C--cycle);
draw(O--A,Arrow(6));
draw(O--B,Arrow(6));
draw(O--C,Arrow(6));

label("$A$", A, A);
label("$B$", B, B);
label("$C$", C, C);
label("$O$", O, NE);
label("$\mathbf{a}$", A/2, SW);
label("$\mathbf{b}$", B/2, SE);
label("$\mathbf{c}$", C/2, SW);
[/asy]

Since $AC = 5,$ $AC^2 = 25.$  But
\begin{align*}
AC^2 &= \|\mathbf{c} - \mathbf{a}\|^2 \\
&= (\mathbf{c} - \mathbf{a}) \cdot (\mathbf{c} - \mathbf{a}) \\
&= \|\mathbf{c}\|^2 - 2 \mathbf{a} \cdot \mathbf{c} + \|\mathbf{a}\|^2 \\
&= 2R^2 - 2 \mathbf{a} \cdot \mathbf{c},
\end{align*}where $R$ is the circumradius.  Hence,
\[\mathbf{a} \cdot \mathbf{c} = R^2 - \frac{AC^2}{2}.\]Similarly, we can prove that
\[\mathbf{a} \cdot \mathbf{b} = R^2 - \frac{AB^2}{2}.\]Therefore,
\[\mathbf{a} \cdot \mathbf{c} - \mathbf{a} \cdot \mathbf{b} = \left( R^2 - \frac{AC^2}{2} \right) - \left( R^2 - \frac{AB^2}{2} \right) = \frac{AB^2 - AC^2}{2} = \frac{3^2 - 5^2}{2} = \boxed{-8}.\]