4. In a Cartesian coordinate system, points $A$, $B$, and $C$ are not collinear. Now, with points $B$ and $C$ as foci, a hyperbola is drawn passing through $A$. With points $A$ and $B$ as foci, a hyperbola is drawn passing through $C$. With points $A$ and $C$ as foci, a hyperbola is drawn passing through $B$. Then, the number of common points of the three hyperbolas is ( ).
(A) 0
(B) 1
(C) 2
(D) Depending on the specific positions of $A$, $B$, and $C$, there are 1 or 2 points
4.D.

Let the first two hyperbolas intersect at point $E$, then
$$
E B - E C = A B - A C, \quad E B - E A = B C - A C.
$$

Subtracting the two equations gives $E C - E A = B C - A B$.
Thus, $E$ lies on the third hyperbola. Therefore, the three hyperbolas intersect at one or two points (since point $E$ may have two positions).