10. In the Cartesian coordinate system, let point $A(0,4)$, $B(3,8)$. If point $P(x, 0)$ makes $\angle A P B$ maximum, then $x=$ $\qquad$
10. $5 \sqrt{2}-3$.

Let $\angle A P B=\theta$.
It is easy to know that when the circumcircle of $\triangle A B P$ is tangent to the $x$-axis, $\angle A P B$ is maximized.
Extend $B A$ to intersect the $x$-axis at point $Q$. Then $Q(-3,0)$. By the power of a point theorem, we have $Q P^{2}=Q A \cdot Q B=5 \times 10$. Thus, $Q P=5 \sqrt{2}$.
Therefore, the x-coordinate of point $P$ at this time is $x=5 \sqrt{2}-3$.