11. (20 points) As shown in Figure 1, in the Cartesian coordinate system $x O y$, the ellipse $\Gamma: \frac{x^{2}}{9}+y^{2}=1$, quadrilateral $A B C D$ is a rhombus, points $A, B, C, D$ are all on the ellipse $\Gamma$, and the diagonals $A C$ and $B D$ intersect at the origin $O$.
(1) Find the minimum area of the rhombus $A B C D$.
(2) Does there exist a fixed circle that is tangent to all rhombi satisfying the given conditions? If it exists, find the equation of the fixed circle; if not, explain the reason.
11. (1) Let points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.

When the diagonals of the rhombus $A B C D$ lie on the coordinate axes, its area is $4 \times \frac{1}{2} \times 3 \times 1=6$;

When the diagonals of the rhombus $A B C D$ do not lie on the coordinate axes, let
$l_{A C}: y=k x$.
Then $l_{B D}: y=-\frac{1}{k} x$.
Solving the system of equations (1) and the equation of the ellipse $\Gamma$ yields
$$
\begin{array}{l}
x_{1}^{2}=\frac{9}{9 k^{2}+1}, y_{1}^{2}=\frac{9 k^{2}}{9 k^{2}+1} \\
\Rightarrow O A^{2}=x_{1}^{2}+y_{1}^{2}=\frac{9\left(k^{2}+1\right)}{9 k^{2}+1} .
\end{array}
$$

Similarly,
$$
\begin{array}{l}
O B^{2}=x_{2}^{2}+y_{2}^{2} \\
=\frac{9\left(\left(-\frac{1}{k}\right)^{2}+1\right)}{9\left(-\frac{1}{k}\right)^{2}+1}=\frac{9\left(k^{2}+1\right)}{k^{2}+9} .
\end{array}
$$

Thus, the area of the rhombus $A B C D$ is
$$
\begin{array}{l}
2OA \cdot O B=18 \sqrt{\frac{k^{4}+2 k^{2}+1}{9 k^{4}+82 k^{2}+9}} \\
=6 \sqrt{\frac{k^{4}+2 k^{2}+1}{k^{4}+\frac{82}{9} k^{2}+1}} \\
=6 \sqrt{1-\frac{\frac{64}{9} k^{2}}{k^{4}+\frac{82}{9} k^{2}+1}} \\
=6 \sqrt{1-\frac{64}{9\left(k^{2}+\frac{1}{k^{2}}\right)+82}} \\
\geqslant 6 \sqrt{1-\frac{64}{9 \times 2 \sqrt{k^{2} \cdot \frac{1}{k^{2}}}+82}} \\
=\frac{18}{5},
\end{array}
$$

with equality holding if and only if $k= \pm 1$.
In summary, the minimum area of the rhombus $A B C D$ is $\frac{18}{5}$.
(2) There exists a fixed circle $x^{2}+y^{2}=\frac{9}{10}$ that is tangent to any rhombus. Let the distance from the origin to any side of the rhombus be $d$.
We will prove that $d=\frac{3}{\sqrt{10}}$.
From (1), we know that when the diagonals of the rhombus $A B C D$ lie on the coordinate axes, $d=\frac{3}{\sqrt{10}}$;
When the diagonals of the rhombus $A B C D$ do not lie on the coordinate axes,
$$
\begin{array}{l}
d^{2}=\frac{O A^{2} \cdot O B^{2}}{O A^{2}+O B^{2}}=\frac{\frac{9\left(k^{2}+1\right)}{9 k^{2}+1} \cdot \frac{9\left(k^{2}+1\right)}{k^{2}+9}}{\frac{9\left(k^{2}+1\right)}{9 k^{2}+1}+\frac{9\left(k^{2}+1\right)}{k^{2}+9}} \\
=\frac{9\left(k^{2}+1\right)^{2}}{\left(k^{2}+1\right)\left(k^{2}+9\right)+\left(k^{2}+1\right)\left(9 k^{2}+1\right)} \\
\left.=\frac{9\left(k^{2}+1\right)^{2}}{\left(k^{2}+1\right)\left(10 k^{2}+10\right)}=\frac{9}{10}, 360227734\right)
\end{array}
$$

Thus, $d=\frac{3}{\sqrt{10}}$.
In summary, there exists a fixed circle $x^{2}+y^{2}=\frac{9}{10}$ that is tangent to all rhombuses.