12. (12 points) Given the line $y=2 x+6$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively. Points $P$ and $Q$ have coordinates $P(-2,0)$ and $Q(0, k)$, where $k<6$. A circle is drawn with $Q$ as the center and $PQ$ as the radius. Then
(1) For what value of $k$ does the circle $\odot Q$ touch the line $AB$;
(2) State the range of values for $k$ when the circle $\odot Q$ is separated from the line $AB$? Intersects the line $AB$? (Only the results need to be provided, no need to show the solution process)
12. Let $x=0$, get $y=6$; let $y=0$, get $x=-3$.
$\therefore$ The coordinates of points $A$ and $B$ are $A(-3,0)$, $B(0,6)$.
$$
\begin{array}{l}
\therefore B Q=6-k, P Q=\sqrt{k^{2}+4}, \\
A B=\sqrt{36+9}=3 \sqrt{5} .
\end{array}
$$

As shown in Figure 5, draw $Q Q_{1} \perp A B$, with the foot of the perpendicular being $Q_{1}$.
Then, from $\mathrm{Rt} \triangle B Q Q$, $\triangle$
$\mathrm{Rt} \triangle E A O$, we get
$$
\frac{B Q}{B A}=\frac{Q Q_{i}}{() A},
$$

i.e., $\frac{B Q}{B A}=\frac{Q P}{B A}$.
$$
\therefore \frac{6 \cdot k}{3 \cdot \sqrt{5}}=\frac{\sqrt{k^{2}+4}}{3} \text {, }
$$

i.e., $k^{2}: j k-4=0$. Solving, we get $k=1$ or -4.
Therefore, when $k=-4$ or $k=1$, $\odot Q$ is tangent to line $A B$;
When $-4<k<1$, $\odot Q$ is separated from line $A B$;

When $1<k<6$ or $k<-4$, $\odot Q$ intersects line $A B$.