Three, there are 10 birds on the ground pecking at food, among which any 5 birds, at least 4 are on a circle. How many birds are there on the circle with the most birds, at minimum?

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Three, there are 10 birds on the ground pecking at food, among which any 5 birds, at least 4 are on a circle. How many birds are there on the circle with the most birds, at minimum?
We use 10 points to represent 10 birds.
(1) Among the 10 points, there must be 5 points that are concyclic. If not, then any 5 points among the 10 points are not concyclic. The 10 known points can form $C_{10}^{5}=252$ groups of 5 points. According to the given, each group of 5 points has four points that are concyclic (hereinafter referred to as a four-point circle), which can form 252 four-point circles (including repeated counts). Each four-point circle belongs to 6 different groups of 5 points, so there should be 42 different four-point circles. Each four-point circle has four known points, totaling 168 points. Since there are 10 known points, there is one point $A$ on 17 different four-point circles. On these 17 four-point circles, apart from $A$, there are 3 other known points, totaling 51 points. Apart from $A$, there are 9 known points, so there is a point $B$ different from $A$, passing through $A$: 17 circles at least 6 pass through $B$, meaning there are at least two different four-point circles passing through $A$ and $B$, which have 12 points and they do not overlap, which is impossible.
(2) Suppose 5 known points $A_{1}, A_{2}, A_{3}, A_{4}$, $A_{8}$ are all on circle $S_{1}$, then the other 5 points can have at most one point not on $S_{1}$. If not, let $B, C$ not be on $S_{1}$.

Consider the 5-point group $A_{1}, A_{2}, A_{3}, B, C$. According to the given, 4 points among the 5 points must be concyclic on $S_{2}$. Since $B, C$ are not on $S_{1}$, $A_{1}, A_{2}, A_{3}$ cannot be on $S_{2}$, otherwise $S_{2}$ and $S_{1}$ would coincide. Assume $A_{1}, A_{2}$ are on $S_{2}$, then $A_{3}, A_{4}, A_{8}$ are not on $S_{2}$.

Consider the 5-point group $A_{3}, A_{4}, A_{5}, B, C$. Assume $A_{3}, A_{4}, B, C$ are concyclic on $S_{3}$, then $A_{1}, A_{2}, A_{5}$ are not on $S_{3}$.
Finally, consider the 5-point group $A_{1}, A_{3}, A_{8}, B, C$. Since $A_{1}, A_{3}, A_{8}$ are on $S_{1}$ and $B, C$ are not on $S_{1}$, $A_{1}, A_{3}, A_{8}, B$ and $A_{1}, A_{3}, A_{8}, C$ are not concyclic. Since $A_{1}, B, C$ are on $S_{2}$ and $A_{3}, A_{5}$ are not on $S_{2}$, $A_{1}, A_{3}, B, C$ and $A_{1}, A_{5}, B, C$ are not concyclic; similarly, it can be proven that $A_{3}, A_{5}, B, C$ are not concyclic, meaning any 4 points in this 5-point group are not concyclic, which contradicts the given.

In summary, the circle containing the most points must have at least 9 known points. Additionally, if 9 known points are on one circle and the other point is not on this circle, these 10 points clearly meet the requirements of the problem. Therefore, the circle with the most birds must have at least 9 birds.