7.55 On a plane, for any 7 points, draw a circle through any 4 of them that are concyclic. How many different circles can be drawn at most?
[Solution] Let $A D, B E, C F$ be the three altitudes of the acute $\triangle A B C$, and $H$ be the orthocenter. Then, a circle can be drawn through 4 points out of the 7 points $A, B, C, D, E, F, H$: $(A, F, H, E), (B, D, H, F), (C, E, H, D), (A, B, D, E), (B, C, E, F), (C, A, F, D)$.

Below, we use proof by contradiction to show that the maximum number of distinct circles that can be drawn is 6. If 7 distinct circles can be drawn, then each of the 7 points must lie on exactly 4 circles. The reasoning is as follows:
(1) At most two circles can pass through two fixed points. If there are 3, then no other points can be common to any two of them. But each circle, apart from the two common points, must have two other known points, which would result in 8 different known points.
(2) At most 4 circles can pass through one fixed point. If there are 5, then because each pair of circles can have at most one other intersection point, and by (1), any 3 of the 5 circles cannot intersect at another point. This would result in at least 9 different known points.
(3) Each circle has 4 known points, and 7 circles have a total of 28 known points (including repeated counts). But by (2), each point can lie on at most 4 circles, so each of the 7 points must lie on exactly 4 circles.

Let the 7 points be $A, B, C, D, E, F, G$, and perform an inversion with $G$ as the center. Then, the 4 circles passing through $G$ become 4 intersecting lines, and the other 3 circles remain as circles. Let the image points be $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}, E^{\prime}, F^{\prime}$. Clearly, for any 4 of these 6 points to be concyclic, no 3 of them can be collinear. Therefore, the 3 four-point circles can only be $\left(A^{\prime}, B^{\prime}, D^{\prime}, F^{\prime}\right), \left(B^{\prime}, C^{\prime}, E^{\prime}, F^{\prime}\right), \left(A^{\prime}, C^{\prime}, D^{\prime}, E^{\prime}\right)$. But since $D^{\prime}$ is inside $\triangle A^{\prime} C^{\prime} E^{\prime}$, these 4 points cannot be concyclic, leading to a contradiction. Thus, the maximum number of distinct circles that can be drawn is 6.