10.41 Given any 5 points in the plane, where no three points are collinear and no four points are concyclic. If a circle passes through three of these points and the other two points are one inside and one outside the circle, it is called a "good circle". If the number of good circles is denoted as $n$, find all possible values of $n$.


[Solution] Choose any two points $A, B$ from 5 points and draw a line through these two points. If the other three points $C, D, E$ are on the same side of line $AB$, then consider $\angle ACB, \angle ADB, \angle AEB$. Without loss of generality, assume $\angle ACB < 180^{\circ}$. Then the circle $ADB$ is the only good circle; if $\angle AEB + \angle ACB < 180^{\circ}$, then the circles $ACB, ADB$, and $AEB$ are all good circles. This means that the number of good circles passing through two fixed points is either 1 or 3.

From 5 points, a total of 10 pairs of points can be formed, and there is at least one good circle passing through each pair of points, so there are at least 10 good circles (including repeated counts). Each good circle passes through exactly 3 pairs of points, so there are at least 4 different good circles, i.e., $n \geq 4$.

Assume $n > 4$. Connect a line segment between every two points among the 5 points, then each line segment is either a chord of a good circle or a common chord of 3 good circles. If there are exactly 5 good circles, then they have a total of 15 chords. Since there are only 10 line segments in total and each line segment contributes 1 or 3 to the count, the sum of 10 odd numbers cannot be 15, so $n \neq 5$. If there are at least 6 good circles, they have at least 18 chords. In this case, the number of line segments contributing 3 to the count is at least 4. Four line segments have 8 endpoints, so there must be two line segments with a common endpoint, say $AB, AC$. Then the circles $ABD, ACD$ are both good circles. This means that the number of good circles passing through $A, D$ is at least 2, and of course, 3. Similarly, the number of good circles passing through $A, E$ is also 3.

Let the shortest of $AB, AC, AD, AE$ be $AB$, then $\angle ACB, \angle ADB$, and $\angle AEB$ are all acute angles. If points $C, D, E$ are on the same side of line $AB$, then the number of good circles passing through $A, B$ is only one, so these 3 points must be on different sides of line $AB$. In this case, since the sum of any two of $\angle ACB, \angle ADB, \angle AEB$ is less than $180^{\circ}$, the number of good circles passing through $A, B$ cannot be 3, which is a contradiction.
In conclusion, the number of good circles $n = 4$.