3. If $A, B, C, D$ are four distinct points on a plane, any circle passing through $A B$ and any circle passing through $C D$ either intersect or coincide, then these four points ( ).
(A) must be collinear
(B) must be concyclic
(C) must be collinear or concyclic
(D) none of the above answers is correct
3. (C).

Let $l_{1}$ and $l_{2}$ be the perpendicular bisectors of $AB$ and $CD$, respectively. If $l_{1} \parallel l_{2}$ or $l_{1}$ coincides with $l_{2}$, then $AB$ and $CD$ are either parallel or collinear. Clearly, $AB \parallel CD$ does not meet the problem's requirements, so $A, B, C, D$ must be collinear, as shown in Figure 1 (interchanging $A$ and $B$ is also acceptable). If $l_{1}$ and $l_{2}$ intersect, the situation shown in Figure 2 is clearly not feasible. Therefore, segments $AB$ and $CD$ must intersect. If $A, B, C, D$ are not concyclic, there exists the scenario shown in Figure 3, which does not meet the problem's requirements. Thus, $A, B, C, D$ must be concyclic. Let $AB$ and $CD$ intersect at $O$, then $AO \cdot OB = CO \cdot OD$. If there exists a circle passing through $AB$ that does not intersect segment $CD$, as shown in Figure 4, extending segment $CD$ in both directions to intersect the circle at $C'$ and $D'$, we have $CO \cdot OD = (C'C + CO) \cdot (DO + DD')$, thus $DD' + CC' + DO \cdot CC' + CO \cdot DD' = 0$, which is a contradiction. Therefore, any circle passing through $AB$ must intersect $CD$ and must pass through points $C$ and $D$, meeting the problem's requirements.