11. (20 points) As shown in Figure 1, given
$\odot G:(x-2)^{2}+y^{2}=r^{2}$ is the incircle of the inscribed $\triangle A B C$ in the ellipse $\frac{x^{2}}{16}+y^{2}=1$, where $A$ is the left vertex of the ellipse.
(1) Find the radius $r$ of $\odot G$;
(2) Two tangents are drawn from point $M(0,1)$ to $\odot G$, intersecting the ellipse at points $E$ and $F$. Prove: The line $E F$ is tangent to $\odot G$.
11. (1) Let point $B\left(2+r, y_{0}\right)$, and draw $G D \perp A B$ at point $D$, and $B C$ intersects the $x$-axis at point $H$.
$$
\begin{array}{l}
\text { By } \frac{G D}{A D}=\frac{H B}{A H} \Rightarrow \frac{r}{\sqrt{36-r^{2}}}=\frac{y_{0}}{6+r} \\
\Rightarrow y_{0}=\frac{r \sqrt{6+r}}{\sqrt{6-r}} .
\end{array}
$$

Since point $B$ is on the ellipse,
$$
\begin{aligned}
y_{0}^{2} & =1-\frac{(2+r)^{2}}{16}=\frac{12-4 r-r^{2}}{16} \\
& =-\frac{(r-2)(r+6)}{16} .
\end{aligned}
$$

From equations (1) and (2), we get $15 r^{2}+8 r-12=0$.
Solving this, we get $r=\frac{2}{3}$ or $r=-\frac{6}{5}$ (discard the latter).
(2) Let the equation of the line passing through point $M(0,1)$ and tangent to the circle $(x-2)^{2}+y^{2}=\frac{4}{9}$ be
$$
\begin{array}{l}
y-1=k x . \\
\text { Then } \frac{2}{3}=\frac{|2 k+1|}{\sqrt{1+k^{2}}} \\
\Rightarrow 32 k^{2}+36 k+5=0 \\
\Rightarrow k_{1}=\frac{-9+\sqrt{41}}{16}, k_{2}=\frac{-9-\sqrt{41}}{16} .
\end{array}
$$

Substituting equation (3) into $\frac{x^{2}}{16}+y^{2}=1$ gives
$$
\left(16 k^{2}+1\right) x^{2}+32 k x=0 \text {. }
$$

Thus, the non-zero solution is $x=-\frac{32 k}{16 k^{2}+1}$.
Let point $F\left(x_{1}, k_{1} x_{1}+1\right), E\left(x_{2}, k_{2} x_{2}+1\right)$.
Then $x_{1}=-\frac{32 k_{1}}{16 k_{1}^{2}+1}, x_{2}=-\frac{32 k_{2}}{16 k_{2}^{2}+1}$.
Thus, the slope of line $E F$ is
$$
\frac{k_{2} x_{2}-k_{1} x_{1}}{x_{2}-x_{1}}=\frac{k_{1}+k_{2}}{1-16 k_{1} k_{2}}=\frac{3}{4} \text {. }
$$

Therefore, the equation of line $E F$ is
$$
y+\frac{32 k_{1}^{2}}{16 k_{1}^{2}+1}-1=\frac{3}{4}\left(x+\frac{32 k_{1}}{16 k_{1}^{2}+1}\right) \text {, }
$$

which simplifies to $y=\frac{3}{4} x-\frac{7}{3}$.
Thus, the distance from the center $(2,0)$ to line $E F$ is $d=\frac{\left|\frac{3}{2}-\frac{7}{3}\right|}{\sqrt{1+\frac{9}{16}}}=\frac{2}{3}$.