5. Given that $A$ and $B$ are two fixed points on a circle $\odot O$ with radius 5, and $P$ is a moving point on $\odot O$. If $AB=6$, let the maximum value of $PA+PB$ be $s$, and the minimum value be $t$, then the value of $s+t$ is $(\quad)$.
(A) $6 \sqrt{10}$
(B) $8 \sqrt{10}$
(C) $6 \sqrt{10}+6$
(D) $6 \sqrt{10}+3$
5.C.

As shown in Figure 5, let the diameter $CD$ of $\odot O$ be perpendicular to the chord $AB$ at point $M$, and let $P$ be any point on $\odot O$ different from point $C$. Draw $CE \perp PB$ at point $E$ and $CF \perp PA$ at point $F$. Then,
$$
\begin{array}{l}
OM=4, CM=9, \\
CA=CB=\sqrt{3^{2}+9^{2}} \\
=3 \sqrt{10}. \\
\text{Also, } \angle CPE=\angle CAB=\angle CBA=\angle CPA, \angle CEP=
\end{array}
$$
$\angle CFP=90^{\circ}$, and $CP$ is a common side, so $\triangle CPE \cong \triangle CPF$, which means $PE=PF$.
Since $AC > AF$ and $BC > BE$, we have
$$
AC + BC > AF + BE = AP + BP.
$$

Therefore, when point $P$ coincides with $C$, $PA + PB$ reaches its maximum value $s = 6\sqrt{10}$; when point $P$ coincides with $A$ or $B$, $PA + PB$ reaches its minimum value $t = AB = 6$.
Thus, $s + t = 6\sqrt{10} + 6$.