17. (12 points) In the Cartesian coordinate system, it is known that $\odot C_{1}$ and $\odot C_{2}$ intersect at points $P$ and $Q$, with the coordinates of $P$ being $(3,2)$. The product of the radii of the two circles is $\frac{13}{2}$. If $\odot C_{1}$ and $\odot C_{2}$ are both tangent to the line $l: y=k x$ and the $x$-axis, find the equation of the line $l$.
17. From the problem, we know that points $O, C_{1}, C_{2}$ are collinear.

Let the radii of $\odot C_{1}$ and $\odot C_{2}$ be $r_{1}$ and $r_{2}$, respectively, and the slope of line $C_{1} C_{2}$ be $\tan \alpha \neq 0$.
Let $m=\cot \alpha$. Then $C_{1}\left(m r_{1}, r_{1}\right), C_{2}\left(m r_{2}, r_{2}\right)$,

The equations of the two circles are
$$
\begin{array}{l}
\left(x-m r_{1}\right)^{2}+\left(y-r_{1}\right)^{2}=r_{1}^{2}, \\
\left(x-m r_{2}\right)^{2}+\left(y-r_{2}\right)^{2}=r_{2}^{2} .
\end{array}
$$

Since $P(3,2)$ is a common point of the two circles, we have
$$
\begin{array}{l}
\left(3-m r_{1}\right)^{2}+\left(2-r_{1}\right)^{2}=r_{1}^{2}, \\
\left(3-m r_{2}\right)^{2}+\left(2-r_{2}\right)^{2}=r_{2}^{2} .
\end{array}
$$

From this, we know that $r_{1}$ and $r_{2}$ are the roots of the equation
$$
m^{2} r^{2}-(6 m+4) r+13=0
$$

Thus, $r_{1} r_{2}=\frac{13}{m^{2}}, m=\sqrt{2}$.
Therefore, the equation of line $l$ is
$$
y=\tan 2 \alpha \cdot x=\frac{2 \tan \alpha}{1-\tan ^{2} \alpha} \cdot x=2 \sqrt{2} x .
$$