11. (20 points) The hyperbola $C: x^{2}-y^{2}=1$ and its symmetric curve $C^{\prime}$ with respect to the line $l: y=k x-1$ have common points. Find the set of real values of $k$.

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11. (20 points) The hyperbola $C: x^{2}-y^{2}=1$ and its symmetric curve $C^{\prime}$ with respect to the line $l: y=k x-1$ have common points. Find the set of real values of $k$.
11. From $\left\{\begin{array}{l}y=k x-1, \\ x^{2}-y^{2}=1\end{array}\right.$, eliminating $y$ yields the quadratic equation in $x$:
$$
\left(1-k^{2}\right) x^{2}+2 k x-2=0 .
$$

When $\Delta=4 k^{2}+8\left(1-k^{2}\right) \geqslant 0$, i.e., $-\sqrt{2} \leqslant k \leqslant \sqrt{2}$, the line $l$ intersects the hyperbola $C$ at some points. These intersection points remain fixed under symmetric transformation, thus the curves $C$ and $C^{\prime}$ have common points.

Suppose there exist two points $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$ on the curve $C$ such that the line $l$ perpendicularly bisects the segment $A B$. In this case, the curves $C$ and $C^{\prime}$ have common points.

Let $l_{A B}: y=-\frac{1}{k} x+b$, and combine it with the equation of the curve $C$, eliminating $x$ yields the quadratic equation in $y$:
$$
\left(k^{2}-1\right) y^{2}-2 k^{2} b y+k^{2} b^{2}-1=0.
$$
This equation has two distinct real roots, thus,
$$
\Delta=4 k^{4} b^{2}-4\left(k^{2}-1\right)\left(k^{2} b^{2}-1\right)>0,
$$

which implies $k^{2}\left(b^{2}+1\right)>1$.
By Vieta's formulas, we have
$$
y_{1}+y_{2}=\frac{2 k^{2} b}{k^{2}-1} \Rightarrow \frac{y_{1}+y_{2}}{2}=\frac{k^{2} b}{k^{2}-1} \text {. }
$$

Notice that,
$$
\begin{array}{l}
\frac{x_{1}+x_{2}}{2}=k\left(b-\frac{y_{1}+y_{2}}{2}\right) \\
=k\left(b-\frac{k^{2} b}{k^{2}-1}\right)=\frac{-k b}{k^{2}-1} .
\end{array}
$$

Thus, the midpoint of segment $A B$ is $E\left(\frac{-k b}{k^{2}-1}, \frac{k^{2} b}{k^{2}-1}\right)$.
Since the line $l: y=k x-1$ passes through point $E$, we have
$$
\begin{aligned}
& \frac{k^{2} b}{k^{2}-1}=\frac{-k^{2} b}{k^{2}-1}-1 \\
\Rightarrow b= & \frac{1-k^{2}}{2 k^{2}} .
\end{aligned}
$$

Substituting (2) into (1) yields
$$
\begin{array}{l}
5 k^{4}-6 k^{2}+1>0 \\
\Rightarrow k \in(-\infty,-1) \cup\left(-\frac{\sqrt{5}}{5}, \frac{\sqrt{5}}{5}\right) \cup(1,+\infty) .
\end{array}
$$

In summary, the range of real numbers $k$ is $\mathbf{R}$.