9. Let the coordinates of point $A$ be $(0,3)$, and points $B, C$ be two moving points on the circle $O: x^{2}+y^{2}=25$, satisfying $\angle B A C=90^{\circ}$. Find the maximum value of the area of $\triangle A B C$.
Given that the midpoint of $BC$ is $D(a, b)$, then
$$
\begin{array}{l}
|OD|^{2}=R^{2}-\left(\frac{1}{2}|BC|\right)^{2}=R^{2}-|AD|^{2} \\
\Rightarrow a^{2}+b^{2}+a^{2}+(b-3)^{2}=25 \\
\Rightarrow a^{2}+\left(b-\frac{3}{2}\right)^{2}=\frac{41}{4} \text{. Thus }|AD|^{2}=a^{2}+(b-3)^{2} \\
=a^{2}+\left(b-\frac{3}{2}\right)^{2}+\frac{27}{4}-3b=17-3b,
\end{array}
$$

When $b=\frac{3-\sqrt{41}}{2}$, the maximum value is obtained, and at this time $AD$ coincides with $AE$.
So $S_{\triangle ABC}=|AD| \cdot|AE| \leqslant|AD|^{2} \leqslant 17-\frac{9-3\sqrt{41}}{2}=\frac{25+3\sqrt{41}}{2}$.
In summary, when $D\left(0, \frac{3-\sqrt{41}}{2}\right)$, the maximum area of $\triangle ABC$ is $\frac{25+3\sqrt{41}}{2}$.