10. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and the moving circle $x^{2}+y^{2}=R^{2}(b<R<a)$, if $A$ is a point on the ellipse, $B$ is a point on the moving circle, and the line $A B$ is tangent to both the ellipse and the moving circle, find the maximum value of the distance $|A B|$ between points $A$ and $B$.
10. Solution: Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, and the equation of line $AB$ be $y=k x+m$. Then the system of equations formed by line $AB$ and the ellipse is
$$
\left\{\begin{array}{l}
y=k x+m \\
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
\end{array} \Rightarrow\left(a^{2} k^{2}+b^{2}\right) x^{2}+2 k m a^{2} x+a^{2}\left(m^{2}-b^{2}\right)=0 .\right.
$$

Since line $AB$ is tangent to the ellipse, we have $\Delta=\left(2 k m a^{2}\right)^{2}-4\left(a^{2} k^{2}+b^{2}\right) a^{2}\left(m^{2}-b^{2}\right)=0$, which gives: $m^{2}=b^{2}+a^{2} k^{2}, x_{1}=-\frac{k m a^{2}}{a^{2} k^{2}+b^{2}}=-\frac{k a^{2}}{m}$.
Similarly, the system of equations formed by line $AB$ and the moving circle is $\left\{\begin{array}{l}y=k x+m \\ x^{2}+y^{2}=R^{2}\end{array}\right.$, which gives: $m^{2}=R^{2}\left(1+k^{2}\right), x_{2}=-\frac{k R^{2}}{m}$. The distance between points $A$ and $B$ on line $AB$ is: $|A B|=\sqrt{1+k^{2}}\left|x_{1}-x_{2}\right|=\sqrt{1+k^{2}}\left|\frac{k\left(R^{2}-a^{2}\right)}{m}\right|$,
Squaring both sides, we get: $|A B|^{2}=\left(1+k^{2}\right) \frac{k^{2}\left(a^{2}-R^{2}\right)^{2}}{m^{2}}=\frac{m^{2}}{R^{2}} \cdot \frac{k^{2}\left(a^{2}-R^{2}\right)^{2}}{m^{2}}=\frac{k^{2}\left(a^{2}-R^{2}\right)^{2}}{R^{2}}$. Since $m^{2}=b^{2}+a^{2} k^{2}, m^{2}=R^{2}\left(1+k^{2}\right)$, we have $k^{2}=\frac{R^{2}-b^{2}}{a^{2}-R^{2}}$, substituting this into the above equation gives
$$
|A B|^{2}=\frac{k^{2}\left(a^{2}-R^{2}\right)^{2}}{R^{2}}=\frac{\left(R^{2}-b^{2}\right)\left(a^{2}-R^{2}\right)}{R^{2}}=a^{2}+b^{2}-\left(R^{2}+\frac{a^{2} b^{2}}{R^{2}}\right) \leqslant(a-b)^{2} \text {, }
$$

with equality holding if and only if $R=\sqrt{a b}$. Therefore, the maximum value of $|A B|$ is $a-b$.