TITLE: The number of co-circular four tuples
QUESTION [5 upvotes]: Let $A,B ⊂ \mathbb{R}$ such that $|A| = |B| = n$. What is the best-known upper bound  on the number of four-tuples in $A \times B$ where the four points are co-circular, they lie on the same circle?

REPLY [2 votes]: Incidence geometry can be used to prove that any finite set $P \subset \mathbb R^2$ with $O(|P|^{1/2})$ points on a single circle determines $O(|P|^3)$ cocircular quadruples. Since your grid point set satisfies this condition, it follows that there are $\ll n^6$ cocircular quadruples. Is this tight?
The argument is similar to that which uses the Szemeredi-Trotter Theorem to show that a set of points in the plane with no more than $|P|^{1/2}$ on a line determine $O(|P|^2 \log |P|)$ collinear triples.
Here is a rough proof. For a fixed point set $P$, let $C_k$ denote the set of all circles containing at least $k$ points from $P$. You can use Pach-Sharir's generalisation of Szemeredi-Trotter for pseudolines to prove that
$$|C_k| \ll \frac{|P|^3}{k^5} + \frac{|P|}{k} \ll \frac{|P|^3}{k^5}, $$
where the last inequality follows from the assumption that $k \ll |P|^{1/2}$. Actually there is a hidden assumption here that $k$ is larger than some absolute constant, but then we can count cocircular quadruples on such poor circles by some trivial arguments.
Now, dyadically decompose the number of cocircular quadruples and apply the bound on $|C_k|$ to complete the proof.
$$|\{ \text{all cocircular quadruples}| \leq \sum_{j \geq 3} \sum_{C : 2^{j-1} \leq | C \cap P| < 2^j} |C \cap P|^4 \ll \sum_j (2^j)^4 \frac{|P|^3}{(2^j)^5} \ll |P|^3.$$