Example 2 Draw two tangents from point $P(-2,-2)$ to circle $\odot M$:
$$
(x-1)^{2}+(y-2)^{2}=9
$$

Let $A$ and $B$ be the points of tangency. Find the equation of the line containing the chord $AB$.
Solution 1: Since the radius $r=3$, the center of the circle is $M(1,2)$, so $|PM|=5$.
Also, $PA$ is a tangent, then $PA \perp AM$. Therefore,
$$
|PA|=\sqrt{|PM|^{2}-|AM|^{2}}=4 \text{. }
$$

Similarly, $|PB|=4$.
Thus, $A$ and $B$ are the intersection points of the circle $(x+2)^{2}+(y+2)^{2}=$ 16 and the circle $(x-1)^{2}+(y-2)^{2}=9$.
Subtracting the two circle equations gives
$$
3 x+4 y-2=0 \text{. }
$$

This is the equation of the chord $AB$ of the tangents.
Comment: The equation
$$
\begin{array}{l}
(x+2)^{2}+(y+2)^{2}-16+ \\
n\left[(x-1)^{2}+(y-2)^{2}-9\right]=0
\end{array}
$$

corresponds to the following curve system:
(1) When $n=-1$, it represents the line passing through the intersection points of the two circles;
(2) When $n \neq-1$, it represents the circle system passing through the intersection points of the two circles.

Solution 2: Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, then the tangent lines at points $A$ and $B$ to the circle
$$
(x-1)^{2}+(y-2)^{2}=9
$$

are respectively
$$
\begin{array}{l}
\left(x_{1}-1\right)(x-1)+\left(y_{1}-2\right)(y-2)=9, \\
\left(x_{2}-1\right)(x-1)+\left(y_{2}-2\right)(y-2)=9 .
\end{array}
$$

Since these two tangent lines both pass through the point $P(-2,-2)$, then
$$
\begin{array}{l}
\left(x_{1}-1\right)(-2-1)+\left(y_{1}-2\right)(-2-2)=9, \\
\left(x_{2}-1\right)(-2-1)+\left(y_{2}-2\right)(-2-2)=9 .
\end{array}
$$

This means that points $A$ and $B$ are both on the line
$$
(x-1)(-2-1)+(y-2)(-2-2)=9
$$

Thus, the equation of the chord $AB$ of the tangents is
$$
3 x+4 y-2=0 \text{. }
$$

Comment: When the point $A\left(x_{1}, y_{1}\right)$ moves on the circle
$$
(x-a)^{2}+(y-b)^{2}=r^{2}
$$

the equation
$$
\left(x_{1}-a\right)(x-a)+\left(y_{1}-b\right)(y-b)=r^{2}
$$

represents the tangent line system at point $A$.