4. $M$ is a moving point on the circle $x^{2}+y^{2}-6 x-8 y=0$, $O$ is the origin, and $N$ is a point on the ray $O M$. If $|O M| \cdot|O N|=150$, then the equation of the locus of point $N$ is $\qquad$ .
$4 \cdot 3 x+4 y-75=0$.
This problem can be solved using the special point method.

As shown in Figure 4, it is easy to know that the radius of the given circle is $r=5$, and the center of the circle is point $O_{1}(3,4)$. Let the ray $O O_{1}$ intersect the circle at $M_{1}$, with $\left|O M_{1}\right|=$ 10. Take point $N_{1}$ on the ray $O O_{1}$ such that $\left|O N_{1}\right|=15$, then
$$
\left|O M_{1}\right| \cdot\left|O N_{1}\right|=150.
$$

Therefore, $N_{1}$ is a point on the trajectory.
Suppose $M_{2}$ is any point on the circle, and point $N_{2}$ is on the ray $O M_{2}$, which is a point on the trajectory, satisfying $\left|O M_{2}\right| \cdot\left|O N_{2}\right|=150$. Thus, $\left|O M_{1}\right| \cdot\left|O N_{1}\right|=\left|O M_{2}\right| \cdot\left|O N_{2}\right|$.

From this, we know that $M_{1}, N_{1}, M_{2}, N_{2}$ are concyclic, and $O M_{1}$ is the diameter, so $\angle O M_{2} M_{1}=90^{\circ}$. Therefore, $\angle M_{1} N_{1} N_{2}=90^{\circ}$.

Hence, the trajectory of point $N$ is a line passing through point $N_{1}(9,12)$ and perpendicular to $O O_{1}$. It is easy to find that the equation of this line is $3 x+4 y-75=0$.