10. (20 points) Given the ellipse $C: \frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, and the moving circle $\Gamma: x^{2}+y^{2}=r^{2}(3<r<5)$. If $M$ is a point on the ellipse $C$, and $N$ is a point on the moving circle $\Gamma$, and the line $M N$ is tangent to both the ellipse $C$ and the moving circle $\Gamma$, find the maximum value of the distance $\mid M N$ ।.
10. Let \( M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right) \),
\[
l_{M N}: y=k x+m \text {. }
\]

Since point \( M \) lies on both the ellipse \( C \) and the line \( M N \), combining the equations of the ellipse and the line and simplifying, we get
\[
\left(25 k^{2}+9\right) x^{2}+50 k m x+25\left(m^{2}-9\right)=0 \text {. }
\]

Since the line \( M N \) is tangent to the ellipse \( C \), we have
\[
\begin{array}{l}
\Delta=(50 k m)^{2}-4\left(25 k^{2}+9\right) \times 25\left(m^{2}-9\right)=0 \\
\Rightarrow m^{2}=9+25 k^{2}, x_{1}=-\frac{25 k}{m} .
\end{array}
\]

Similarly, since point \( N \) lies on both the circle \( \Gamma \) and the line \( M N \), we get
\[
m^{2}=r^{2}\left(1+k^{2}\right), x_{2}=-\frac{k r^{2}}{m} .
\]

From conclusions (1) and (2), we have
\[
\begin{array}{l}
k^{2}=\frac{r^{2}-9}{25-r^{2}}, x_{2}-x_{1}=\frac{k\left(25-r^{2}\right)}{m} . \\
\text { Hence }|M N|^{2}=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2} \\
=\left(1+k^{2}\right)\left(x_{2}-x_{1}\right)^{2} \\
=\frac{m^{2}}{r^{2}} \cdot \frac{k^{2}\left(25-r^{2}\right)^{2}}{m^{2}} \\
=\frac{\left(25-r^{2}\right)^{2}}{r^{2}} \cdot \frac{r^{2}-9}{25-r^{2}} \\
=\frac{\left(25-r^{2}\right)\left(r^{2}-9\right)}{r^{2}} \\
=25+9-r^{2}-\frac{25 \times 9}{r^{2}} \\
\leqslant 4,
\end{array}
\]

with equality holding if and only if \( r=\sqrt{5 \times 3} \). Therefore, the maximum value of \( |M N| \) is 2.