11. (20 points) Let the moving line $l$ be tangent to the circle $\odot O: x^{2}+y^{2}=1$, and intersect the ellipse $\frac{x^{2}}{9}+y^{2}=1$ at two distinct points $A$ and $B$. Find the maximum distance from the origin to the perpendicular bisector of $AB$.

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II. 11. According to the problem, let $l: y=k x+m(k \neq 0)$.
Since the line $l$ is tangent to $\odot O$, the distance from the origin $O$ to the line $l$ is 1, i.e.,
$$
\frac{|m|}{\sqrt{1+k^{2}}}=1 \text {. }
$$

Let points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. By combining the equations of the ellipse and the line, we get
$$
\begin{array}{l}
\left(1+9 k^{2}\right) x^{2}+18 k m x+\left(9 m^{2}-9\right)=0 \\
\Rightarrow x_{1}+x_{2}=-\frac{18 k m}{1+9 k^{2}} .
\end{array}
$$

Thus, the equation of the perpendicular bisector of $A B$ is
$$
y-\frac{m}{1+9 k^{2}}=-\frac{1}{k}\left(x+\frac{9 k m}{1+9 k^{2}}\right) \text {, }
$$

i.e., $x+k y+\frac{8 k m}{1+9 k^{2}}=0$.
Therefore, the distance from the origin $O$ to it is
$$
d=\frac{\left|\frac{8 k m}{1+9 k^{2}}\right|}{\sqrt{1+k^{2}}}=\frac{|8 k|}{1+9 k^{2}} \leqslant \frac{4}{3},
$$

with equality holding if and only if $|k|=\frac{1}{3}$.