If we replace $10$ with $n\ge 5$, then I believe the answer is $n-1$ for $n\ge 7$ and $4$ for $n=5,6$. The case $n=5$ is obvious, and the fact that $4$ works when $n=6$ can be seen by considering the $6$ vertices of two unit squares in the plane which share a side. From now on we assume $n\ge 7$. First of all, we notice that if we can find a circle with $5$ points, then that circle must contain $n-1$ points. Suppose the contrary, i.e. that we can find a circle with at least five points, $A,B,C,D,E$, and two other points, $X,Y$, not belonging to that circle. In the group $X,Y,A,B,C$ we must have four concyclic points, and one of $A,B,C$ must not belong to this group, so we may assume WLOG the group is $X,Y,A,B$. Now look at the group $X,Y,A,C,D$. The group of $4$ concyclic points that we can extract from must contain $X,Y$, so it cannot contain $A$. This means that it's $X,Y,C,D$. Finally, look at the group $X,Y,A,C,E$. The group of four concyclic points extracted from here must contain $X,Y$, but cannot contain either $A$ or $C$, so we get a contradiction. What we need to do now is find those $n$ for which there is a configuration satisfying the hypothesis for which every circle passing through at least four points passes through no other points except for those four. We will show that no $n\ge 7$ works. Obviously, in order to do this, it suffices to prove it for $n=7$. Let $A,B,C,D$ be four concyclic points in our configuration. Just as above, if $X,Y,Z$ are the other three points, then we can find, say, circles through the quadruples $(X,Y,A,B),(X,Y,C,D)$ and $(X,Z,A,C),(X,Z,B,D)$. Now perform an inversion of pole $X$. For a general point $P$, we denote its image by $P'$. $A',B',C',D'$ are still concyclic, and $Y'=A'B'\cap C'D',\ Z'=A'C'\cap B'D'$. The six points $A',B',C',D',Y',Z'$ must still satisfy the hypothesis, but it's clear that no four of the points $Y',Z',A',C',D'$ are concyclic.