7.83 On the ground, there are 10 birds pecking at food, among which any 5 birds, at least 4 are on a circle. How many birds are on the circle with the most birds, at minimum?
[Solution 1] Use 10 points to represent 10 birds. If any 4 points among the 10 points are concyclic, then the 10 points are all on the same circle. Below, let points $A, B, C, D$ not be concyclic. At this time, a circle can be drawn through any 3 non-collinear points among the 4 points, and at most 4 different circles $S_{1}, S_{2}, S_{3}, S_{4}$ can be drawn, and at least 3 different circles can be drawn. For these two cases, the following argument is consistent, we only prove the case of 4 different circles.

Take any point $P_{i}$ from the remaining 6 points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}, P_{6}$ and form a five-point group with $A, B, C, D$. According to the given, there must be 4 points concyclic among them. Therefore, point $P_{i}$ must be on one of the circles $S_{1}, S_{2}, S_{3}, S_{4}$. By the arbitrariness of $P_{i}$, each of the 6 points must fall on one of the 4 circles. By the pigeonhole principle, there must be two points on the same circle, i.e., there must be 5 points concyclic among the 10 points.
Assume $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ are on the same circle $C_{1}$, and points $P$ and $Q$ are not on $C_{1}$.
(1) Consider the five-point group $\left\{A_{1}, A_{2}, A_{3}, P, Q\right\}$, among which there must be 4 points concyclic on $C_{2}$. $C_{2}$ must contain at least one of $P$ and $Q$, so $C_{2} \neq C_{1}$, hence $A_{1}, A_{2}, A_{3}$ cannot all be on $C_{2}$, otherwise $C_{2}$ would coincide with $C_{1}$. Without loss of generality, assume $A_{1}, A_{2}, P, Q \in C_{2}$, then $A_{3}, A_{4}, A_{5} \notin C_{2}$.
(2) Consider the five-point group $\left\{A_{3}, A_{4}, A_{5}, P, Q\right\}$, among which there must be 4 points concyclic on $C_{3}$. Clearly, $C_{3} \neq C_{1}$, so we can assume $A_{3}, A_{4}, P, Q \in C_{3}$, then $A_{1}, A_{2}, A_{5} \notin C_{3}$. It is evident that $C_{3} \neq C_{2}$.
(3) Consider the five-point group $\left\{A_{1}, A_{3}, A_{5}, P, Q\right\}$, among which there must be 4 points concyclic on $C_{4} \neq C_{1}$. Therefore, $A_{1}, A_{3}, A_{5}$ cannot all be on $C_{4}$, so $P, Q \in C_{4}$ and at least one of $A_{1}$ and $A_{3}$ belongs to $C_{4}$.

If $A_{1} \in C_{4}$, then $C_{4}$ coincides with $C_{2}$. But since $A_{3}, A_{5} \notin C_{2}$ and one of them belongs to $C_{4}$, this is a contradiction. If $A_{3} \in C_{4}$, then $C_{4}$ coincides with $C_{3}$. But since $A_{1}, A_{5} \notin C_{3}$, they do not belong to $C_{4}$, which is impossible.

In summary, we have proven that there is at most one point among the 10 points outside the circle $C_{1}$, i.e., $C_{1}$ has at least 9 points. On the other hand, the scenario where 9 points among the 10 known points are on the same circle and the other 1 point is not on this circle clearly meets the requirements of the problem, so the circle with the most birds has at least 9 birds.
[Solution 2] We use 10 points to represent 10 birds and first prove that there must be 5 points concyclic among the 10 points.

If not, then any 5 points among the 10 points are not concyclic, but there must be 4 points concyclic, which we call a four-point circle. The 10 known points can form $C_{10}^{5}=252$ five-point groups, each of which can form a four-point circle, totaling 252 four-point circles (including repeated counts). Each four-point circle belongs to 6 different five-point groups, so there are 42 different four-point circles.
The 42 four-point circles have a total of 168 known points, and there are 10 different known points, so by the pigeonhole principle, there is a point $A$ such that the number of four-point circles passing through point $A$ is at least 17.

On the 17 four-point circles passing through point $A$, there are 3 known points on each circle besides point $A$, totaling 51 known points. These 51 points are the other 9 known points besides $A$, so by the pigeonhole principle, there is another point $B \neq A$ such that at least 6 of the 17 four-point circles pass through point $B$. This means that there are at least 6 four-point circles passing through points $A$ and $B$.

On these 6 four-point circles, there are 2 known points on each circle besides points $A$ and $B$, totaling 12 points, which are the other 8 known points besides $A$ and $B$. By the pigeonhole principle, there is another point $C$ such that at least 2 of the 6 circles pass through point $C$. Thus, these two different four-point circles have 3 common points $A, B, C$. Therefore, the two circles coincide, leading to 5 points being concyclic, which contradicts the assumption. This proves that there must be 5 points concyclic among the 10 points.
The following proof is the same as Solution 1.
[Solution 3] Suppose 4 points $A, B, C, D$ among the 10 known points are concyclic. Perform an inversion transformation with point $A$ as the center, then the images of points $B, C, D$ are $B^{\prime}, C^{\prime}, D^{\prime}$ on a straight line, and together with the images of the other 6 points, the 9 points, any 4 points are either concyclic or 3 points are collinear.

For simplicity, we will still use the original letters to denote the images of points $B, C, D, \cdots, I, J$. If there are two points $E, F$ outside the line $B C$, then consider 3 four-point groups $\{E, F, B, C\},\{E, F, B, D\},\{E, F, C, D\}$. Clearly, exactly one of these groups is concyclic. Without loss of generality, assume $E, F, C, D$ are concyclic. Then $B, E, F$ are collinear. Consider $B, C, F$ and the image of the 7th point $G$, then point $G$ is on the circle $B C F$ or on the line of one of the sides of $\triangle B C F$.
(1) If $G$ is on the line $B C$, then $G, D, E, F$ 4 points are neither concyclic nor any 3 points are collinear, which is a contradiction. Therefore, $G$ is not on the line $B C$. Similarly, $G$ is not on the line $B F$.
(2) If $G$ is on the line $C F$, then $G$ is not on the circle $D E F$, so $G$ must be on the line $D E$, i.e., $G$ is the intersection of lines $C F$ and $D E$. Clearly, there is at most 1 such intersection (see the figure above).
(3) If point $G$ is on the circle $B C F$, then $G$ is not on the circle $C D F$, nor on the line $C D$ or $C F$, so $G$ must be on the line $D F$. Thus, $G$ is the other intersection of line $D F$ and circle $B C F$, i.e., $G^{\prime}$ in the figure above. There is also at most 1 such intersection.

The above derivation for point $G$ applies equally to the images of the other 4 points $G, H, I, J$, so these 4 points must be one of the two intersections, which is impossible. This proves that there is at most 1 image point outside the line $B C$. Therefore, the original circle $A B C D$ has at least 9 known points.

On the other hand, the scenario where 9 points among the 10 known points are on the same circle and the 10th point is not on this circle clearly meets the requirements of the problem, so the circle with the most birds has at least 9 birds.