Example 3 There are ten birds on the ground pecking at food, and among any five birds, at least four birds are on the same circumference. Question: What is the minimum number of birds on the circumference that contains the most birds?
For convenience, let a red dot represent a bird, and adopt the "inch by inch" approach. First, we prove that there must be a circle with five points.

In fact, if there is no circle with five points, ten points can form $\mathrm{C}_{10}^{5}=252$ "five-point groups," and each five-point group has a four-point circle, resulting in 252 four-point circles.
Next, consider the repetition.
Since there are six points outside each four-point circle, each four-point circle belongs to six five-point groups (by adding one point to a four-point circle, we get a five-point group), meaning it repeats six times.

Therefore, there are at least $\frac{252}{6}=42$ different four-point circles, and each four-point circle has four red dots, resulting in $4 \times 42 = 168$ red dots.

Since there are only 10 red dots in total, there must be a red dot $A$ that belongs to at least 17 circles. In these 17 circles, apart from the common point $A$, there are three other red dots, totaling $3 \times 17 = 51$ red dots.

Among the 10 different red dots, excluding point $A$, there are only nine points, so there must be a point $B$ different from point $A$ that appears at least $\left[\frac{51-1}{9}\right]+1=6$ times (where $[x]$ represents the greatest integer not exceeding the real number $x$), meaning point $B$ belongs to at least six circles. In these six circles, apart from the common points $A$ and $B$, there are two other red dots, totaling 12 red dots, and they are all different, which is impossible. Therefore, there must be a circle with five points.

Now, assume five red dots $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ are all on circle $S_{1}$. We only need to prove that among the remaining five red dots, at most one point is not on circle $S_{1}$ (i.e., circle $S_{1}$ has at least nine red dots).
If not, assume there are two points $B$ and $C$ not on circle $S_{1}$.
Consider the five-point group $B, C, A_{1}, A_{2}, A_{3}$.
According to the condition, there is a four-point circle $S_{2}$ among them.
Obviously, points $A_{1}, A_{2}, A_{3}$ cannot all be on circle $S_{2}$, otherwise, circle $S_{1}$ will coincide with circle $S_{2}$, contradicting the fact that "points $B$ and $C$ are not on circle $S_{1}$."
Assume $A_{3} \notin S_{2}$, and $B, C, A_{1}, A_{2} \in S_{2}$.
Then, points $A_{3}, A_{4}, A_{5}$ are not on circle $S_{2}$ (if any of them were on circle $S_{2}$, circle $S_{1}$ would coincide with circle $S_{2}$, contradicting the fact that "points $B$ and $C$ are not on circle $S_{1}$.")
Next, consider the five-point group $B, C, A_{3}, A_{4}, A_{5}$.
Assume $B, C, A_{3}, A_{4}$ are on circle $S_{3}$. Then, points $A_{1}, A_{2}, A_{5}$ are not on circle $S_{3}$.
Finally, summarizing the above results:
$$
\begin{array}{l}
S_{1}=\left(A_{1}, A_{2}, A_{3}, A_{4}, A_{5}\right), B, C \notin S_{1} ; \\
S_{2}=\left(B, C, A_{1}, A_{2}\right), A_{3}, A_{4}, A_{5} \notin S_{2} ; \\
S_{3}=\left(B, C, A_{3}, A_{4}\right), A_{1}, A_{2}, A_{5} \notin S_{3} .
\end{array}
$$

Consider the five-point group $B, C, A_{1}, A_{3}, A_{5}$, which must have a four-point circle $S_{4}$, and the four-point groups in this five-point group are only five cases:
$$
\begin{array}{l}
\left(B, C, A_{1}, A_{3}\right),\left(B, C, A_{1}, A_{5}\right), \\
\left(B, C, A_{3}, A_{5}\right),\left(B, A_{1}, A_{3}, A_{5}\right), \\
\left(C, A_{1}, A_{3}, A_{5}\right) .
\end{array}
$$

If $S_{4}=\left(B, C, A_{1}, A_{3}\right)$, then $A_{3} \in S_{2}$, which is a contradiction;
If $S_{4}=\left(B, C, A_{1}, A_{5}\right)$, then $A_{5} \in S_{2}$, which is a contradiction;
If $S_{4}=\left(B, C, A_{3}, A_{5}\right)$, then $A_{5} \in S_{3}$, which is a contradiction;
If $S_{4}=\left(B, A_{1}, A_{3}, A_{5}\right)$, then $B \in S_{1}$, which is a contradiction;
If $S_{4}=\left(C, A_{1}, A_{3}, A_{5}\right)$, then $C \in S_{1}$, which is a contradiction.
Therefore, in the five-point group $B, C, A_{1}, A_{3}, A_{5}$, any four points do not lie on the same circle, contradicting the condition.
Thus, the assumption is false, meaning that among the remaining five points in
$$
S_{1}=\left(A_{1}, A_{2}, A_{3}, A_{4}, A_{5}\right)
$$

at most one point is not on circle $S_{1}$, indicating that circle $S_{1}$ has at least nine red dots.

On the other hand, if nine points are on the same circle and the other point is not on this circle, these ten points clearly satisfy the problem's requirements.

Therefore, the circle with the most birds has at least nine birds.