10. Let the distances from $P$ to two vertices $A, B$ of the equilateral $\triangle A B C$ be $2, 3$ respectively. Then the maximum value that $P C$ can reach is ( ).
(A) $\sqrt{5}$
(B) $\sqrt{13}$
(C) 5
(D)6
10.C.

As shown in Figure 7, rotate $AB$ so that $AB$ coincides with $AC$, and $P$ rotates to $P^{\prime}$. Therefore,
$$
\begin{array}{r}
\angle P A B = \angle P^{\prime} A C . \\
\text{Thus, } \angle P A P^{\prime} = 60^{\circ} . \\
\text{Hence, } P P^{\prime} = P A = P^{\prime} A \\
= 2, P^{\prime} C = P B = 3 \text{. Therefore, }
\end{array}
$$
$$
\text{Thus, } P C \leqslant P P^{\prime} + P^{\prime} C = 5 \text{. }
$$