Example 5: On the ground, there are 10 birds pecking, and among any 5 birds, at least 4 birds are on the same circumference. What is the maximum number of birds on the circumference that has the most birds? (6th China Mathematical Olympiad)
Analysis and Solution: Use points to represent birds. Let the circle with the most points have $r$ points. Clearly, $r \geqslant 4$. Can $r=4$? Let's first explore the case where $r=4$.

If $r=4$, i.e., each circle has at most 4 points, but we are considering circles that pass through at least 4 points, so each "4-point circle" has exactly 4 points.

Step 1: Calculate the number of "4-point circles". Note the condition: any 5-point set has 4 points on a circle, i.e., each "5-point set" corresponds to a "4-point circle". Thus, we can use a mapping to count the number of circles. In fact, each "5-point set" corresponds to one "4-point circle", so there are $C_{10}^{5}=252$ "4-point circles". However, each "4-point circle" can belong to 6 different "5-point sets", and is counted 6 times, so the number of "4-point circles" is $\frac{252}{6}=42$. These "4-point circles" are distinct. If not, there would be two different 4-point sets $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ on the same circle, but $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ would have at least 5 distinct points, and these 5 points would be on the same circle, contradicting $r=4$.

Step 2: Observe the partition and use a middle quantity to count twice. The 42 different 4-point circles can be seen as 42 different (possibly intersecting) subsets, denoted as: $M_{1}, M_{2}, \cdots, M_{42}$, which satisfy: $\left|M_{i}\right|=4$, $\left|M_{i} \cap M_{j}\right| \leqslant 2$. Using the "addition technique", we can calculate the number of triangles $S$.

On one hand, $S=C_{10}^{3}=120$. On the other hand, $S \geqslant 42 \mathrm{C}_{4}^{3}=168$, which is a contradiction. Therefore, $r>4$.
Let $M$ be the circle with the most points, and by $r>4$, $M$ has at least 5 points $A, B, C, D, E$. We will prove: the other points (except possibly one point) are on this circle. (*)

In fact, suppose there are two points $P, Q$ not on circle $M$, then among the 5 points $P, Q, A, B, C$, 4 points are on a circle. But $P, Q$ are not on circle $A B C$, so $P, Q$ must be on a circle with two of $A, B, C$, say $P, Q, A, B$ are on circle $M_{1}$.

Similarly, consider the 5 points $P, Q, C, D, E$, there must be 4 points among $P, Q$ and $C, D, E$ on a circle, say $P, Q, C, D$ are on circle $M_{2}$. Now consider the 5 points $P, Q, A, C, E$, there must be 4 points among $P, Q$ and $A, C, E$ on a circle.
(i) If $P, Q, A, C$ are on circle $M_{3}$, then $M_{3}$ coincides with $M_{1}$, so $P Q A B C$ are on the same circle, and $P, Q$ are on circle $M$, a contradiction.
(ii) If $P, Q, A, E$ are on circle $M_{3}$, then $M_{3}$ coincides with $M_{1}$, so $P Q A B E$ are on the same circle, and $P, Q$ are on circle $M$, a contradiction.
(iii) If $P, Q, C, E$ are on circle $M_{3}$, then $M_{3}$ coincides with $M_{2}$, so $P Q C D E$ are on the same circle, and $P, Q$ are on circle $M$, a contradiction.

In summary, conclusion $(*)$ holds, i.e., $r \geqslant 9$. Finally, $r=9$ is possible, i.e., 9 points are on the same circle, and the other point is outside the circle, which clearly meets the condition.

In conclusion, the minimum value of $r$ is 9.