Existence of lattices whose circles have bounded number of points

Question

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?

Answer 1

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.

Answer 2

Let $p(m,n)$ be the point $mA+nB.$ Even if you allow $\Lambda$ to include all such points as $m,n$ range over the rationals (rather than the integers), it is easy to have no circle include $4$ points of $\Lambda.$ Let $A=(1,0)$ and $B=(\pi,y)$ for some fixed but not yet determined $y>0.$ Any line containing two points of $\Lambda$ contains infinitely many, but these are not circles. Given four points $p(m_1,n_1),p(m_2,n_2),p(m_3,n_3),p(m_4,n_4)$ there is a unique circle containing the first three whose center depends on $y$ (provided that they are not on a common line.) For the fourth point $p(m_4,n_4)$ to be on the same circle reduces to an equation $ay^2=b$ which has at most one positive solution. There are only a countable number of $4$-tuples of points and an uncountable number of choices for $y.$

If we took $B=(0,y)$ then there would be $4$-tuples making a rectangle and hence on a common circle, but it is easy to avoid $5$ on a circle.