5. In the Cartesian coordinate system, circle $C_{1}$ intersects with circle $C_{2}$ at points $P$ and $Q$, where the coordinates of point $P$ are $(3,2)$, and the product of the radii of the two circles is $\frac{13}{2}$. If the line $y=k x(k>0)$ and the $x$-axis are both tangent to circles $C_{1}$ and $C_{2}$, then $k=$ $\qquad$
5. $2 \sqrt{2}$.

Let the coordinates of the centers of the two circles be $C_{i}\left(x_{i}, y_{i}\right)(i=1,2)$. Then $\left|P C_{i}\right|=y_{i}$, that is, $\sqrt{\left(x_{i}-3\right)^{2}+\left(y_{i}-2\right)^{2}}=y_{i}$, which also means $x_{i}^{2}-6 x_{i}-4 y_{i}+13=0(i=1,2)$.
The line connecting the centers of the two circles $C_{1} C_{2}$ passes through the origin $O$, and is the angle bisector of the line $y=k x$ and the $x$-axis. Let its equation be $y=m x$. Then
$$
y_{i}=m x_{i}(i=1,2) .
$$

Substituting equation (2) into equation (1) yields
$$
x_{i}^{2}-(6+4 m) x_{i}+13=0 \text {. }
$$

Therefore, $x_{1} x_{2}=13$.
Also, from the problem, we know $y_{1} y_{2}=\frac{13}{2}$, that is, $m^{2} x_{1} x_{2}=\frac{13}{2}$. Hence $m^{2}=\frac{1}{2}$, which means $m=\frac{\sqrt{2}}{2}$ (since equation (3) has no solution, $-\frac{\sqrt{2}}{2}$ is discarded). Using the double-angle tangent formula, we get $k=\frac{2 m}{1-m^{2}}=2 \sqrt{2}$.