3. In a Cartesian coordinate system, two circles have a common point $(9,6)$ and are both tangent to the $x$-axis. The product of their radii is 68. If their other external common tangent also passes through the origin, then its slope is $\qquad$ .
3. $\frac{12 \sqrt{221}}{49}$.

Since the line connecting the centers of the two circles passes through the origin, we can set the centers of the two circles as $(a, k a),(b, k b)$. Thus,
$$
\begin{array}{l}
(a-9)^{2}+(k a-6)^{2}=(k a)^{2} \\
\Rightarrow a^{2}-6(2 k+3) a+117=0 .
\end{array}
$$

Similarly, $b^{2}-6(2 k+3) b+117=0$.
Since $a \neq b$, $a$ and $b$ are the two roots of the quadratic equation
$$
x^{2}-6(2 k+3) x+117=0
$$

Thus, $a b=117$.
Since $k a \cdot k b=68$, we have $k=\sqrt{\frac{68}{117}}=\frac{2 \sqrt{17}}{3 \sqrt{13}}$.
The slope of the other external common tangent is twice the inclination angle of the line connecting the centers, so the slope is $\frac{2 k}{1-k^{2}}=\frac{12 \sqrt{221}}{49}$.