11. (15 points) Let the system of inequalities $\left\{\begin{array}{l}x+y>0, \\ x-y<0\end{array}\right.$ represent the plane region $D$. The product of the distances from a moving point $P$ in region $D$ to the lines $x+y=0$ and $x-y=0$ is 2. The trajectory of point $P$ is denoted as curve $C$. A line $l$ passing through the point $F(2 \sqrt{2}, 0)$ intersects curve $C$ at points $A$ and $B$. If the circle with diameter $A B$ is tangent to the $y$-axis, find the slope of line $l$.
11. Solution: From the problem, we know that the plane region $D$ is as shown by the shaded area in the figure. Let the moving point be $P(x, y)$, then $\frac{|x+y|}{\sqrt{2}} \cdot \frac{|x-y|}{\sqrt{2}}=2$, which means $\left|x^{2}-y^{2}\right|=4$. Since $P \in D$, we have $x+y>0, x-y>0$, which leads to the equation of the curve $C$ as $\frac{y^{2}}{4} - \frac{x^{2}}{4}=1(y>0)$.

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then the center of the circle with diameter $A B$ is $Q\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$. Since the circle $L$ with diameter $A B$ is tangent to the $y$-axis, the radius $r=\frac{1}{2}|A B| = \left|\frac{x_{1}+x_{2}}{2}\right|$, which means $|A B|=\left|x_{1}+x_{2}\right|$.

Since the line $A B$ passes through the point $F(2 \sqrt{2}, 0)$, when $A B \perp x$-axis, it does not meet the problem's condition.
Therefore, let the equation of the line $A B$ be $y=k(x-2 \sqrt{2})$.
Substituting into the hyperbola equation $\frac{y^{2}}{4}-\frac{x^{2}}{4}=1(y>0)$, we get
$k^{2}(x-2 \sqrt{2})^{2}-x^{2}=4$, which simplifies to $\left(k^{2}-1\right) x^{2}-4 \sqrt{2} k^{2} x + \left(8 k^{2}-4\right)=0$.
Since the line intersects the hyperbola at points $A, B$,
we have $k \neq \pm 1$.
Thus, $x_{1}+x_{2}=\frac{4 \sqrt{2} k^{2}}{k^{2}-1}, x_{1} x_{2}=\frac{8 k^{2}-4}{k^{2}-1}$.
Therefore, $|A B|=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}} = \sqrt{\left(1+k^{2}\right)\left[\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}\right]}$
$=\sqrt{\left(1+k^{2}\right)\left[\left(\frac{4 \sqrt{2} k^{2}}{k^{2}-1}\right)^{2}-4 \cdot \frac{8 k^{2}-4}{k^{2}-1}\right]} = \left|x_{1}+x_{2}\right| = \left|\frac{4 \sqrt{2} k^{2}}{k^{2}-1}\right|$,
Simplifying, we get: $k^{4}+2 k^{2}-1=0$,
Solving, we get $k^{2}=\sqrt{2}-1\left(k^{2}=-\sqrt{2}-1\right.$ is not valid, discard).
By $\Delta=\left(4 \sqrt{2} k^{2}\right)^{2}-4\left(k^{2}-1\right)\left(8 k^{2}-4\right)=3 k^{2}-1>0$,
and since $y>0$, we have $-1<k<-\frac{\sqrt{3}}{3}$. Therefore, $k = -\sqrt{\sqrt{2}-1}$.