5. For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>b>0)$, the left focus is $F_{1}$, the vertices are $A_{1}$ and $A_{2}$, and $P$ is any point on the hyperbola. Then the two circles with diameters $P F_{1}$ and $A_{1} A_{2}$ respectively must ( ).
(A) intersect
(B) be tangent
(C) be separate
(D) any of the above situations is possible
5.B.

As shown in Figure 1, let point $P$ be on the right branch of the hyperbola. Taking the midpoint $M$ of $P F_{1}$, then $|O M|$ is the distance between the centers of the two circles, $\frac{|P F_{1}|}{2}$ and $a$ are the radii of the two circles, respectively. In $\triangle P F_{1} F_{2}$, $|O M|=\frac{1}{2}|P F_{2}|$.

Also, $|P F_{1}|-|P F_{2}|=2a$, which means $|P F_{1}|=|P F_{2}|+2a$, thus $|O M|=\frac{|P F_{1}|}{2}-a$.

It can be seen that the circle with $P F_{1}$ as its diameter is internally tangent to the circle with $A_{1} A_{2}$ as its diameter.

If point $P$ is on the left branch of the hyperbola, it is easy to know that $\frac{|P F_{1}|}{2}$ and $a$ are the radii of the two circles, respectively.
Since $|P F_{2}|-|P F_{1}|=2a$, then
$|O M|=\frac{1}{2}|P F_{2}|$, which means $|O M|=\frac{|P F_{1}|}{2}+a$.
This indicates that the circle with $P F_{1}$ as its diameter is externally tangent to the circle with $A_{1} A_{2}$ as its diameter.