Problem 1699

There are at most two such points. For example,
the points $(0,0)$ and $(1,0)$ lie on a circle with center
$(1/2, x)$ for any real number $x$, not necessarily rational.

On the other hand, suppose $P = (a,b), Q = (c,d), R = (e,f)$
are three rational points that lie
on a circle. The midpoint $M$ of the side $PQ$ is
$((a+c)/2, (b+d)/2)$, which is again rational. Moreover, the slope
of the line $PQ$ is $(d-b)/(c-a)$, so the slope of the line through
$M$ perpendicular to $PQ$ is $(a-c)/(b-d)$, which is rational or infinite.

Similarly, if $N$ is the midpoint of $QR$, then $N$ is a rational point
and the line through $N$ perpendicular to $QR$ has rational slope.
The center of the circle lies on both of these lines, so its
coordinates $(g,h)$ satisfy two linear equations with rational
coefficients, say $Ag + Bh = C$ and $Dg + Eh = F$. Moreover,
these equations have a unique solution. That solution must then be
$$\begin{align*} g &= (CE - BD)/(AE - BD) \\ h &= (AF - BC)/(AE - BD) \end{align*}$$
(by elementary algebra, or Cramer's rule),
so the center of the circle is rational. This proves the desired result.