13. (25 points) As shown in Figure 2, in the Cartesian coordinate system $x O y$, let $M\left(x_{0}, y_{0}\right)$ be a point on the ellipse $C: \frac{x^{2}}{4}+y^{2}=1$, with the left and right foci being $F_{1}$ and $F_{2}$, respectively. Draw two tangents from the origin $O$ to the circle $\odot M:\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}=r^{2}(0<r<1)$, which intersect the ellipse $C$ at points $P$ and $Q$. The slopes of the lines $O P$ and $O Q$ are denoted as $k_{1}$ and $k_{2}$, respectively.
(1) Let the lines $M F_{1}$ and $M F_{2}$ intersect the circle $\odot M$ at points $A$ and $B$, respectively. When $\left|A F_{1}\right|-\left|B F_{2}\right|=2 r$, find the equation of the locus of point $A$;
(2) When $k_{1} k_{2}$ is a constant, find the maximum value of $|O P||O Q|$.
13. (1) By the definition of an ellipse,
$$
\begin{array}{l}
\left|M F_{1}\right|+\left|M F_{2}\right|=2 a \\
\Rightarrow 2 r+\left|A F_{1}\right|+\left|B F_{2}\right|=2 a \\
\Rightarrow\left|A F_{1}\right|+\left|B F_{2}\right|=2 a-2 r . \\
\text { Also, }\left|A F_{1}\right|-\left|B F_{2}\right|=2 r, \text { then } \\
\left|A F_{1}\right|=a=2 .
\end{array}
$$

Thus, the coordinates of point $A(x, y)$ satisfy the equation
$$
(x+\sqrt{3})^{2}+y^{2}=4 \text {. }
$$

By $00$ or $x0) \text {. }
$$
(2) Let the equation of line $O P$ be $y=k_{1} x$, which is tangent to $\odot M$, then
$$
\frac{\left|k_{1} x_{0}-y_{0}\right|}{\sqrt{k_{1}^{2}+1}}=r,
$$

which gives $\left(x_{0}^{2}-r^{2}\right) k_{1}^{2}-2 x_{0} y_{0} k_{1}+y_{0}^{2}-r^{2}=0$.
Similarly, the line $O Q$ is tangent to $\odot M$, and let the equation of line $O Q$ be $y=k_{2} x$, then
$$
\left(x_{0}^{2}-r^{2}\right) k_{2}^{2}-2 x_{0} y_{0} k_{2}+y_{0}^{2}-r^{2}=0 \text {. }
$$

Thus, $k_{1} 、 k_{2}$ are the two real roots of the equation
$$
\left(x_{0}^{2}-r^{2}\right) x^{2}-2 x_{0} y_{0} x+y_{0}^{2}-r^{2}=0
$$

Hence, $k_{1} k_{2}=\frac{y_{0}^{2}-r^{2}}{x_{0}^{2}-r^{2}}=\frac{1-\frac{x_{0}^{2}}{4}-r^{2}}{x_{0}^{2}-r^{2}}$.
Since $k_{1} k_{2}$ is a constant, let $k_{1} k_{2}=c$, then
$$
\begin{aligned}
k_{1} k_{2} & =\frac{1-\frac{x_{0}^{2}}{4}-r^{2}}{x_{0}^{2}-r^{2}}=c \\
& \Rightarrow\left(c+\frac{1}{4}\right) x_{0}^{2}=1+(c-1) r^{2} .
\end{aligned}
$$

Since $M$ is a point on the ellipse and $c$ is a constant, it must be that $c=-\frac{1}{4}$.
Thus, $k_{1} k_{2}=-\frac{1}{4}$, and at this time, $r=\frac{2 \sqrt{5}}{5}$.
Let the points $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$.
From $\left\{\begin{array}{l}y=k_{1} x, \\ \frac{x^{2}}{4}+y^{2}=1\end{array}\right.$
$$
\Rightarrow x_{1}^{2}=\frac{4}{1+4 k_{1}^{2}}, y_{1}^{2}=\frac{4 k_{1}^{2}}{1+4 k_{1}^{2}} \text {. }
$$

Similarly, $x_{2}^{2}=\frac{4}{1+4 k_{2}^{2}}, y_{2}^{2}=\frac{4 k_{2}^{2}}{1+4 k_{2}^{2}}$.
Thus, $O P^{2} \cdot O Q^{2}$
$$
\begin{array}{l}
=\left(\frac{4}{1+4 k_{1}^{2}}+\frac{4 k_{1}^{2}}{1+4 k_{1}^{2}}\right)\left(\frac{4}{1+4 k_{2}^{2}}+\frac{4 k_{2}^{2}}{1+4 k_{2}^{2}}\right) \\
=\frac{4\left(1+k_{1}^{2}\right)}{1+4 k_{1}^{2}} \cdot \frac{4\left(1+k_{2}^{2}\right)}{1+4 k_{2}^{2}} \\
=\frac{4\left(1+k_{1}^{2}\right)}{1+4 k_{1}^{2}} \cdot \frac{1+16 k_{1}^{2}}{1+4 k_{1}^{2}} \\
\leqslant \frac{\left(\frac{5+20 k_{1}^{2}}{2}\right)^{2}}{\left(1+4 k_{1}^{2}\right)^{2}}=\frac{25}{4} .
\end{array}
$$