Does there exist a circle with EXACTLY two rational points, no more, no less

Question

Topic title says it all, I think. Does there exist a circle with EXACTLY two rational points, no more, no less? I know that as soon as it has three rational points, its center must be rational too, therefore an isomorphism with the unit circle yields infintely many (dense) rational points. I have examples with only one and zero rational points, but I'm wondering about the case 2.

I already know that the center must be irrational, because else one rational point on the circle would suffice to yield infinitely many points.

Answer 1

From cut-the-knot.org:

there are at most two rational points on a circle with irrational center. Before proving the result, let's recollect a few fundamental facts.

For any pair of rational points, the straight line passing through the two may be expressed by an equation with rational (and, hence, integer) coefficients. The midpoint of the segment joining two rational points is itself a rational point. A line perpendicular to a line with slope $\alpha$ has a slope $-1/\alpha$, implying that the slopes of the two lines are either both rational or both irrational. Therefore, the perpendicular bisector of a segment joining two rational points always has an equation with rational coefficients. The solution to a system of two linear equations with rational coefficients, if exists, is necessarily rational. Which is to say, that if two straight lines that are expressed by linear equations with rational coefficients intersect, their point of intersection is necessarily rational.

With these preliminaries, we are ready to confront the problem. Assume to the contrary there is a circle with irrational center on which there are three distinct rational points, say, $P, Q, R$. The perpendicular bisectors of $PQ$ and $QR$ would meet at the center of the circle which, from the foregoing discussion, would be a rational point in contradiction with the conditions of the problem.

So, choose two rational points $P,Q$ and an irrational point $C$ on their perpendicular bisector. The circle with center $C$ that goes through $P,Q$ answers your question.