Inspired by Marta Reece

Suppose that 3 or more rational points lie on it. Thus the circle is the circumcircle of the triangle formed by 3 of those points. By coordinate geometry, the centre is the intersection of the perpendicular bisectors, which is found by solving 2 linear equations. Thus the center is rational, a contradiction.

As pointed out by Zee Ell, I should provide a construction for completeness.

One such construction is the circle centred at $(\sqrt2,\sqrt2)$ with radius $\sqrt{5-2\sqrt2}$ . This passes through $(0,1)$ and $(1,0)$ .

The area of a circle with a radius of $r$ is $r^2\pi$ , so $r^2\pi>\left \lfloor r^2\pi \right \rfloor$ . Since the Blichfeldt's Theorem it can be translated, such that it covers minimum $\left \lfloor r^2\pi \right \rfloor +1$ grid points.

If we give a positive integer, let this number be $N$ , then there exist a circle on the grid, which covers exactly $N$ points. This is because: Let the origo be $O(\sqrt{2}; \sqrt{3})$ . Now the circle can't contain more than 1 point, because if there were, then the perpendicular to the midpoint of the segment between the points equation would be $ax+by=c$ , where $a, b, c$ are integers, and $a\sqrt{2}+b\sqrt{3}=c$ would have to be true, but squaring it: $\sqrt{6}=\dfrac{c^2-2a^2-3b^2}{2ab}$ which is impossible, since $\sqrt{6}$ is irrational.

Now write a small circle over $O$ , which doesn't cover any point, and then increase the radius constantly. The growing circle will only go through 1 point, so we can increase the circle, such that it covers exactly $N$ points.

With this continuous increase it is also possible to add a circle on the square grid that does not have a grid point on its perimeter, but its radius is arbitrarily large.

But if we write a big circle around a grid point, it is only a little easier to avoid grid points, the more true it will be:

The distance of a circle around a grid point from its nearest grid point can be tapped down if the radius of the circle is sufficiently large (János Surányi)