5. Given two points $A(3,1), B(-3,-5)$ in the Cartesian coordinate system $x O y$, find a point $P$ on the $x$-axis such that the value of $|P A-P B|$ is maximized. Then the maximum value of $|P A-P B|$ is ( ).
(A) $4 \sqrt{2}$
(B) $5 \sqrt{2}$
(C) $2 \sqrt{13}$
(D) $5 \sqrt{3}$
5.C.

As shown in Figure 4, construct the point $A_{1}(3,-1)$, which is the reflection of point $A$ over the $x$-axis.

The line $A_{1} B$ intersects the $x$-axis at $P$. Then point $P$ maximizes the value of $\mid P A - P B \mid$.

Assume that $P_{1}$ is any point on the $x$-axis different from point $P$, and connect $P_{1} A$, $P_{1} A_{1}$, $P_{1} B$, and $P A$.
Since $A$ and $A_{1}$ are symmetric with respect to the $x$-axis, we have
$$
P A = P A_{1}, \quad P_{1} A = P_{1} A_{1}.
$$

In $\triangle P_{1} A_{1} B$, we have $\left|P_{1} A_{1} - P_{1} B\right| < A_{1} B$.
Thus, $\left|P_{1} A - P_{1} B\right| = \left|P_{1} A_{1} - P_{1} B\right| < A_{1} B$
$$
= \left|P B - P A_{1}\right| = |P B - P A| = |P A - P B|.
$$

Therefore, point $P$ maximizes the value of $|P A - P B|$, and the maximum value is the length of segment $A_{1} B$, which is
$$
A_{1} B = \sqrt{(3 + 3)^{2} + (-1 + 5)^{2}} = 2 \sqrt{13}.
$$