Consider the $n^2$ lattice points $(i, j)$, where $1 \leq i, j \leq n$.
Let the number of circles that pass through at least 3 points of this set be $C(n)$. What is a good way to count this? Is there a good way to count this? Trivially, there are at most ${n \choose 3}$ such circles. Note that as $n$ gets large, you can have several points that lie on the same circle.
I'm tempted to set up a recurrence relation. By considering the lower left and upper right $n-1$ by $n-1$ square, we can show $C(n) = 2 C(n-1) - C(n-2) +$ some weird additional terms.
Edit: I believe $C(3) = 34$. 14 of the circles contain 4 points, and 20 contain 3 points. 8 sets of 3 points form a line, for a total of $14 \times 4 + 20 + 8 = 84 = { 9 \choose 3 }$ set of 3 points which can form a circle.
Edit: It was slightly surprising when I first found out that $(1, 1), (2,1), (3, 2), (3, 3)$ lie on the same circle. (This follows by symmetry, with center of circle $( 1.5, 2.5)$ ).
I do not know how to approach $C(4)$