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.
