9. (16 points) Let point $A(0,3)$, and two moving points $B, C$ on the circle $\odot O: x^{2}+y^{2}=$ 25 satisfy $\angle B A C=90^{\circ}$. Find the maximum value of the area of $\triangle A B C$.

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The translation maintains the original format and line breaks as requested.
Given $B\left(x_{1}, y_{1}\right), C\left(x_{2}, y_{2}\right)$, and $P(x, y)$ as the midpoint of line segment $B C$.
$$
\begin{array}{l}
\text { Then } x_{1}^{2}+y_{1}^{2}=25, x_{2}^{2}+y_{2}^{2}=25, \\
x_{1} x_{2}+\left(y_{1}-3\right)\left(y_{2}-3\right)=0, \\
x_{1}+x_{2}=2 x, y_{1}+y_{2}=2 y .
\end{array}
$$

From the above equations, we know
$$
\begin{array}{l}
x^{2}+y^{2}-3 y=8 \\
\Rightarrow x^{2}+\left(y-\frac{3}{2}\right)^{2}=\left(\frac{\sqrt{41}}{2}\right)^{2} .
\end{array}
$$

Thus, the trajectory of the midpoint $P$ of line segment $B C$ is $\odot O_{1}$, with the equation
$$
x^{2}+\left(y-\frac{3}{2}\right)^{2}=\left(\frac{\sqrt{41}}{2}\right)^{2} \text {. }
$$

Therefore, $|A P| \leqslant\left|A O_{1}\right|+\left|O_{1} P\right|=\frac{3}{2}+\frac{\sqrt{41}}{2}$.
Then $S_{\triangle A B C}=\frac{1}{2}|A B||A C|$
$$
\begin{array}{l}
\leqslant \frac{1}{4}\left(|A B|^{2}+|A C|^{2}\right)=\frac{1}{4}|B C|^{2}=|A P|^{2} \\
\leqslant\left(\frac{3}{2}+\frac{\sqrt{41}}{2}\right)^{2}=\frac{25+3 \sqrt{41}}{2},
\end{array}
$$

When point $P\left(0, \frac{3-\sqrt{41}}{2}\right)$, the equality holds.
Hence, the maximum area of $\triangle A B C$ is $\frac{25+3 \sqrt{41}}{2}$.