10 In the Cartesian coordinate system, it is known that points $A(0,4)$ and $B(3,8)$. If point $P(x, 0)$ makes $\angle APB$ the largest, then $x=$ $\qquad$.
(10) Let $\angle A P B=\theta$. Since the circle with diameter $A B$ does not intersect the $x$-axis, $\theta<\frac{\pi}{2}$. To maximize $\angle A P B$, we need $\sin \theta$ to be maximized, which in turn requires the circumradius of $\triangle A B P$ to be minimized. It is easy to see that the circumradius of $\triangle A B P$ is minimized when its circumcircle is tangent to the $x$-axis. Extend $B A$ to intersect the $x$-axis at point $Q$, then $Q(-3,0)$. By the power of a point theorem, $Q P^{2}=Q A \cdot Q B=5 \cdot 10$. Therefore, $Q P=5 \sqrt{2}$, and at this point, the $x$-coordinate of point $P$ is $x=5 \sqrt{2}-3$.