16. Given the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and the moving circle $\Gamma: x^{2}+y^{2}=R^{2}$, where $b<R<a$. If $A$ is a point on the ellipse $E$, and $B$ is a point on the moving circle $\Gamma$, and the line $A B$ is tangent to both the ellipse $E$ and the moving circle $\Gamma$, find the maximum value of the distance $|A B|$ between points $A$ and $B$.
16. Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, and the equation of line $AB$ be $y=k x+m$.
Since $A$ is on both the ellipse $E$ and the line $AB$, we have
$$
\left\{\begin{array}{l}
y_{1}=k x_{1}+m, \\
\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=1 .
\end{array}\right.
$$

Substituting (1) into (2) gives
$$
\left(a^{2} k^{2}+b^{2}\right) x^{2}+2 k m a^{2} x+a^{2}\left(m^{2}-b^{2}\right)=0 \text {. }
$$

Since the line $AB$ is tangent to the ellipse $E$, we have
$$
\Delta=\left(2 k m a^{2}\right)^{2}-4\left(a^{2} k^{2}+b^{2}\right) a^{2}\left(m^{2}-b^{2}\right)=0 \text {. }
$$

Thus, we get
$$
m^{2}=b^{2}+a^{2} k^{2}, x_{1}=-\frac{k a^{2}}{m} \text {. }
$$

Similarly, since $B$ is on both the circle $\Gamma$ and the line $AB$, we get
$$
m^{2}=R^{2}\left(1+k^{2}\right), x_{2}=-\frac{k R^{2}}{m} \text {. }
$$

From (3) and (4), we have
$$
\begin{array}{l}
k^{2}=\frac{R^{2}-b^{2}}{a^{2}-R^{2}}, x_{2}-x_{1}=\frac{k\left(a^{2}-R^{2}\right)}{m} . \\
\text { Hence }|A B|^{2}=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2} \\
=\left(1+k^{2}\right)\left(x_{2}-x_{1}\right)^{2} \\
=\frac{m^{2}}{R^{2}} \cdot \frac{k^{2}\left(a^{2}-R^{2}\right)^{2}}{m^{2}}=\frac{\left(a^{2}-R^{2}\right)^{2}}{R^{2}} \cdot \frac{R^{2}-b^{2}}{a^{2}-R^{2}} \\
=\frac{\left(a^{2}-R^{2}\right)\left(R^{2}-b^{2}\right)}{R^{2}}=a^{2}+b^{2}-R^{2}-\frac{a^{2} b^{2}}{R^{2}} \\
=(a-b)^{2}-\left(R-\frac{a b}{R}\right)^{2} \leqslant(a-b)^{2} .
\end{array}
$$

Equality holds if and only if $R=\sqrt{a b}$. 
Therefore, the maximum value of the distance $|A B|$ between points $A$ and $B$ is $a-b$.