10. (15 points) Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, $M(m, 0)(0<m<a)$ is a moving point. Let $f(m)$ be the radius of the smallest circle centered at point $M$ that is tangent to the ellipse $C$. Find $f(m)$.
10. Since $\odot M$ is the circle with the smallest radius that is tangent to the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, the radius should be the minimum distance from a point on the ellipse to $M$. Let any point on the ellipse be $P\left(x_{0}, y_{0}\right)$. Then $\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}=1$. Therefore,
$$
\begin{array}{l}
M P^{2}=\left(x_{0}-m\right)^{2}+y_{0}^{2}=\left(x_{0}-m\right)^{2}+\left(1-\frac{x_{0}^{2}}{a^{2}}\right) b^{2} \\
=\left(1-\frac{b^{2}}{a^{2}}\right) x_{0}^{2}-2 m x_{0}+m^{2}+b^{2} \\
=\left(1-\frac{b^{2}}{a^{2}}\right)\left(x_{0}-\frac{m}{1-\frac{b^{2}}{a^{2}}}\right)^{2}-\frac{m^{2}}{1-\frac{b^{2}}{a^{2}}}+m^{2}+b^{2} \\
=\left(1-\frac{b^{2}}{a^{2}}\right)\left(x_{0}-\frac{m a^{2}}{a^{2}-b^{2}}\right)^{2}-\frac{m^{2} b^{2}}{a^{2}-b^{2}}+b^{2} . \\
\text { And } \frac{m a^{2}}{a^{2}-b^{2}} \leqslant a \Leftrightarrow m \leqslant \frac{a^{2}-b^{2}}{a}=\frac{c^{2}}{a}, \text { thus, }
\end{array}
$$

When $m \leqslant \frac{c^{2}}{a}$, the minimum value of $M P$ is
$$
\begin{array}{l}
\sqrt{-\frac{m^{2} b^{2}}{a^{2}-b^{2}}+b^{2}}=\sqrt{-\frac{m^{2} b^{2}}{c^{2}}+b^{2}} \\
=\sqrt{\frac{b^{2}\left(c^{2}-m^{2}\right)}{c^{2}}}=\frac{b}{c} \sqrt{c^{2}-m^{2}} ;
\end{array}
$$

When $m>\frac{c^{2}}{a}$, the minimum value of $M P$ is obtained when $x_{0}=a$.
Thus, the minimum value of $M P$ is $a-m$.
In summary, $f(m)=\left\{\begin{array}{ll}\frac{b}{c} \sqrt{c^{2}-m^{2}}, & 0<m \leqslant \frac{c^{2}}{a} ; \\ a-m, & \frac{c^{2}}{a}<m<a .\end{array}\right.$