11. The vertex $C(x, y)$ of $\triangle A B C$ satisfies the inequalities $x^{2}+y^{2} \leqslant 8+2 y, y \geqslant 3$, and side $A B$ lies on the $x$-axis. Given that the point $Q(0,1)$ is at a distance of 1 from the lines $A C$ and $B C$, the minimum area of $\triangle A B C$ is $\qquad$.
11. $6 \sqrt{2}$

From $x^{2}+y^{2} \leqslant 8+2 y$, we know $x^{2}+(y-1)^{2} \leqslant 9$, and also $y \geqslant 3$, so point $C$ is within the segment of a circle with center $Q(0,1)$ and radius 3. At the same time, the circle with center $Q$ and radius 1 is the incircle of $\triangle A B C$. Let $A\left(a_{1}, 0\right), B\left(a_{2}, 0\right), C\left(x_{0}, y_{0}\right)$, the equation of line $A C$ is: $(x -a_{1}) y_{0}-y\left(x_{0}-a_{1}\right)=0$, the distance from point $(0,1)$ to line $A C$ is 1, i.e., $d=$ $\frac{\left|x_{0}-a_{1}-a_{1} y_{0}\right|}{\sqrt{\left(x_{0}-a_{1}\right)^{2}+y_{0}^{2}}}=1$,
simplifying gives
$$
\left(y_{0}-2\right) a_{1}^{2}+2 x_{0} a_{1}-y_{0}=0
$$

Similarly,
$$
\left(y_{0}-2\right) a_{2}^{2}+2 x_{0} a_{2}-y_{0}=0
$$

Thus, $a_{1}, a_{2}$ are the two distinct roots of the equation $\left(y_{0}-2\right) a^{2}+2 x_{0} a-y_{0}=0$. According to Vieta's formulas,
$|A B|^{2}=\left|a_{1}-a_{2}\right|^{2}=\left(a_{1}+a_{2}\right)^{2}-4 a_{1} a_{2}=\frac{4\left[x_{0}^{2}-y_{0}\left(y_{0}-2\right)\right]}{\left(y_{0}-2\right)^{2}}$
Here $y_{0} \in[3,4]$.
When $y_{0}$ is determined, to maximize $|A B|$, by (1), $x^{2}$ should take its maximum value $9-\left(y_{0}-1\right)^{2}$, so when $y_{0} \in[3,4]$ and is a certain value, point $C$ should be on the arc of the segment to maximize $|A B|$, thus the area of $\triangle A B C$ also reaches its maximum.

When $y_{0}$ is on the arc of the segment, $x_{0}^{2}+y_{0}\left(y_{0}-2\right)=9-\left(y_{0}-1\right)^{2}+y_{0}\left(y_{0}-2\right)=8$, then $|A B|=\frac{4 \sqrt{2}}{y_{0}-2}$.
Therefore, $S_{\triangle A B C}=\frac{1}{2} y_{0} \frac{4 \sqrt{2}}{y_{0}-2}=2 \sqrt{2}\left(1+\frac{2}{y_{0}-2}\right)$.
Clearly, when $y_{0}=3$, the area of $\triangle A B C$ reaches its maximum value $6 \sqrt{2}$, at this time point $C$ is at the midpoint of the segment's chord.