6 Given that point $P$ is a moving point on the line $l: k x+y+4=0(k>0)$, $P A$ and $P B$ are two tangent lines to the circle $C: x^{2}+y^{2}-2 y=0$, and $A$ and $B$ are the points of tangency. If the minimum area of quadrilateral $P A C B$ is 2, then $k=$ $\qquad$.
(6) 2 Hint: Since
$$
\begin{aligned}
S_{\text {quadrilateral } P A C B} & =|P A| \cdot|A C|=|P A| \\
& =\sqrt{|C P|^{2}-|C A|^{2}}=\sqrt{|C P|^{2}-1},
\end{aligned}
$$

Therefore, the area is minimized when $|C P|$ is minimized, i.e., when $C P \perp l$.
From $\sqrt{|C P|^{2}-1}=2$ we get $|C P|=\sqrt{5}$, and by the point-to-line distance formula we have
$$
|C P|=\sqrt{5}=\frac{5}{\sqrt{1+k^{2}}},
$$

Since $k>0$, we get $k=2$.