For any plane lattice $\Lambda= \{ mA+nB: m,n \in \mathbb Z \}$, with $A,B$ linearly independent vectors in $\mathbb R^2$, we define the set of the circles in $\Lambda$ as


$$\mathcal K(\Lambda) = \Bigl\{\bigl\{X \in \Lambda:\|X-C\|=R\bigr\} : C \in \mathbb R^2 , R \in \mathbb R\_{\ge 0} \Bigr\},$$


where $\|\cdot\|$ is the usual Euclidean metric. Let's define the function $\psi\_\Lambda$, for every plane lattice $\Lambda$,
\begin{align}
\psi\_{\Lambda}\colon \mathcal K(\Lambda) &\to \mathbb N\\
\gamma &\mapsto |\gamma|.
\end{align}



> 
> *Conjecture.* There is a plane lattice $\Lambda$ such that $\psi\_\Lambda$ is bounded.
> 
> 
> 


It is not difficult to find lattices $\Lambda$ whose circles with centres in $\Lambda$ have at most two points. But, is my conjecture true in the general case?



Existence of lattices whose circles have bounded number of points
I think, for most lattices even no four points are concyclic. Indeed, consider a lattice generated by complex numbers 1 and $z$. Assume that four points 0, $a\_1+b\_1z$, $a\_2+b\_2z$, $a\_3+b\_3z$ with integer $a\_1,a\_2,a\_3,b\_1,b\_2,b\_3$ lie on a circle. Then the cross-ratio $$\frac{a\_1+b\_1z}{a\_2+b\_2z}:\frac{a\_1-a\_3+(b\_1-b\_3)z}{a\_2-a\_3+(b\_2-b\_3)z}$$
must be real. This is an algebraic condition on the real and imaginary parts of $z$, which is nontrivial unless we are speaking about four points on a line (but a line is not a circle). The plane can not be covered by countably many algebraic curves, thus there exist $z$ such that no 4 points lie on a circle.


