6. Let $k$ be a real number, in the Cartesian coordinate system there are two point sets $A=\left\{(x, y) \mid x^{2}+y^{2}=2(x+y)\right\}$ and $B=\{(x, y) \mid k x-y+k+3 \geq 0\}$, if $A \cap B$ is a singleton set, then the value of $k$ is $\qquad$
Answer: $-2-\sqrt{3}$. Solution: The point set $\mathrm{A}$ is the circumference of the circle $\Gamma:(x-1)^{2}+(y-1)^{2}=2$, and the point set $\mathrm{B}$ is the line $l: y-3=k(x+1)$ passing through the point $\mathrm{P}(-1,3)$ and its lower part (including the boundary). By plotting these two point sets, we know that when $\mathrm{A}$ and $\mathrm{B}$ are a single set, the line $l$ is a tangent to the circle $\Gamma$ passing through point $\mathrm{P}$. Therefore, the distance from the center of the circle $\Gamma$, $\mathrm{M}(1,1)$, to the line $l$ equals the radius of the circle $\sqrt{2}$, so $\frac{|k-1+k+3|}{\sqrt{k^{2}+1}}=\sqrt{2}$. Combining with the graph, the smaller root $k=-2-\sqrt{3}$ should be taken.