$$
\begin{array}{l}
\text { 11. Given } A=\left\{(x, y) \mid x^{2}+y^{2}-2 x \cos \alpha+\right. \\
2(1+\sin \alpha)(1-y)=0, \alpha \in \mathbf{R}\}, \\
B=\{(x, y) \mid y=k x+3, k \in \mathbf{R}\} .
\end{array}
$$

If $A \cap B$ is a singleton set, then $k=$ $\qquad$
$$
\begin{array}{l}
\text { 11. } \pm \sqrt{3} \text {. } \\
x^{2}+y^{2}-2 x \cos \alpha+2(1+\sin \alpha)(1-y)=0 \\
\Rightarrow(x-\cos \alpha)^{2}+(y-1-\sin \alpha)^{2}=0 \\
\Rightarrow x=\cos \alpha, y=1+\sin \alpha \\
\Rightarrow x^{2}+(y-1)^{2}=1 .
\end{array}
$$

Notice that $\boldsymbol{A} \cap B$ is a singleton set, meaning the line $y=k x+3$ is tangent to $x^{2}+(y-1)^{2}=1$, then $k= \pm \sqrt{3}$.