Example 8 Given the line $y=2 x+6$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively, and points $P$ and $Q$ have coordinates $P(-2,0)$ and $Q(0, k)$, where $k<6$. A circle is drawn with point $Q$ 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)
(1999, Huanggang City, Hubei Province Junior High School Mathematics Competition)
Solution: (1) From the given, we have $A(-3,0)$, $B(0, 6)$. Therefore,
$$
\begin{array}{l}
B Q=6-k, \\
P Q=\sqrt{k^{2}+4}, \\
A B=3 \sqrt{5} .
\end{array}
$$

As shown in Figure 5, draw $Q R \perp A B$ at $R$, then
Rt $\triangle B Q R \backsim$ Rt $\triangle B A O$, so
$$
\frac{B Q}{A B}=\frac{Q R}{O A} \text {. }
$$

Since $Q R=P Q$,
we have $\frac{B Q}{A B}=\frac{P Q}{O A}$. Thus, $\frac{6-k}{3 \sqrt{5}}=\frac{\sqrt{k^{2}+4}}{3}$, which simplifies to $k^{2}+3 k-4=0$.
Solving this, we get $k=1$ or $-4$.
Therefore, when $k=-4$ or $k=1$, $\odot Q$ is tangent to the line $A B$.
(2) When $-4<k<1$, $\odot Q$ is separated from the line $A B$; when $1<k<6$ or $k<-4$, $\odot Q$ intersects the line $A B$.