11. Given a moving line $l$ that is tangent to the circle $O: x^{2}+y^{2}=1$ and intersects the ellipse $\frac{x^{2}}{9}+y^{2}=1$ at two distinct points $A, B$, find the maximum distance from the origin to the perpendicular bisector of $A B$.
According to the problem, we can set $l: y=k x+m(k \neq 0)$.
Since the line $l$ is tangent to the circle $O$, the distance from $O$ to the line $l$ is 1, i.e., $\frac{|m|}{\sqrt{1+k^{2}}}=1$.
(5 points) Such a line must intersect the ellipse at two different points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. By solving the system $\left\{\begin{array}{l}y=k x+m \\ x^{2}+9 y^{2}-9=0\end{array}\right.$, we get
$$
\left(1+9 k^{2}\right) x^{2}+18 k m x+\left(9 m^{2}-9\right)=0
$$

We obtain $x_{1}+x_{2}=-\frac{18 k m}{1+9 k^{2}}$, so the coordinates of the midpoint of $A B$ are $\left(-\frac{9 k m}{1+9 k^{2}}, \frac{m}{1+9 k^{2}}\right)$.
The equation of the perpendicular bisector of $A B$ is $y-\frac{m}{1+9 k^{2}}=-\frac{1}{k}\left(x+\frac{9 k m}{1+9 k^{2}}\right)$, which simplifies to $x+k y+\frac{8 k m}{1+9 k^{2}}=0$.
The distance from $O$ to the perpendicular bisector of $A B$ is $d=\frac{\left|\frac{8 k m}{1+9 k^{2}}\right|}{\sqrt{1+k^{2}}}$.
Substituting $\frac{|m|}{\sqrt{1+k^{2}}}=1$ into $d=\frac{\left|\frac{8 k m}{1+9 k^{2}}\right|}{\sqrt{1+k^{2}}}$, we get $d=\frac{|8 k|}{1+9 k^{2}}$. By the AM-GM inequality, $1+9 k^{2} \geq 6|k|$, so $d \leq \frac{4}{3}$, with equality if and only if $|k|=\frac{1}{3}$. Therefore, when $|k|=\frac{1}{3},|m|=\frac{\sqrt{10}}{3}$, the maximum distance from the origin to the perpendicular bisector of $A B$ is $\frac{4}{3}$.
(20 points)