So this is a problem that really got me thinking,
We have a circle with a center at $(\sqrt2,0)$ we need to look for points on the circle with rational coordinates, the radius is rational.
We are not given any radius, so what I did was use parametric form to write any point on the circle
Let $(x,y)$ be any point on the circle of radius $r$, we can write
$$x=\sqrt2 + r\cos{\theta},~~y=r\sin\theta.$$
When I solve for rational values I get only two or one such points exist.
I don't have a official answer as such, a friend of mine gave answer as $3$, three points exist. Not really sure what the correct answer is?
Best Answer
(Correcting my previous (incorrect) answer.)
Suppose there is such a circle given by $(x-\sqrt 2)^2+y^2=r^2$. If $x,y$, and $r$ are all rational, then since $x^2-2\sqrt 2x+2+y^2=r^2$, we see that $2\sqrt 2 x=x^2+y^2+2-r^2$ is rational. If $x\neq 0$, then $\sqrt 2=\frac{xy^2+2-r^2}{2x}$ is rational, which is impossible. Thus $x=0$, so $y=\pm\sqrt{r^2-2}$.
Let $r=p/q$ for some integers $p,q$ with $q>0$. In order for $y$ to be rational, we must have
$$\sqrt{\frac{p^2}{q^2}-2}=\sqrt{\frac{p^2-2q^2}{q^2}}=\frac1q\sqrt{p^2-2q^2}\in\Bbb Q.$$
Since $p^2-2q^2$ is an integer, $\sqrt{p^2-2q^2}$ is rational if and only if $p^2-2q^2$ is a perfect square. Thus, there are at most two points on the circle $(x-2\sqrt 2)^2+y^2=r^2$ with rational coordinates, and finding such a circle is equivalent to finding integers $p$ and $q$ for which $p^2-2q^2$ is a perfect square. I can't answer this in general, but $p=3$ and $q=2$ gives one instance.
In conclusion, a circle centered at $(\sqrt 2,0)$ has either $0$ or $2$ rational points, an example of the latter being $(x-\sqrt 2)^2+y^2=\left(\frac32\right)^2$ with $\left(0,\pm\frac12\right)$.