Rationals on circles

Here my proof that the answer is $\boxed{2}$ .

First I will prove a property on the radius $R$ if there are rational points.

Suppose there is a rational point on the circle, so:

$R\cos(\theta)=r_1 \in\mathbb{Q}$

and,

$R\sin(\theta)+\pi=\pm \sqrt{R^2-r_1^2}+\pi=r_2\in\mathbb{Q}$

ie, $\pm \sqrt{R^2-r_1^2}=r_2-\pi$

squaring this equality,

$R^2=\pi^2 + r_1^2+r_2^2-2r_2\pi \ (*)$

Therefore, $R$ has to be in this form:

$R=\sqrt{\pi^2+q_1\pi+q_2}$ with $q_1, \ q_2$ rationals and $q_2\geq 0$ .

Plugging in into $(*)$ ,

$q_1\pi + q_2 = r_1^2+r_2^2-2r_2\pi \\ ie, \ (q_1+2r_2)\pi=r_1^2+r_2^2-q_2$

So because $\pi$ is irrational, $r_2$ has to be $-\frac{q_1}{2}$ and then $r_1^2$ has to be $q_2-r_2^2=\frac{4q_2-q_1^2}{4}$ . In fact it is a proof that the function $R: (q_1,q_2)\in\mathbb{Q}\times\mathbb{Q}_+ \mapsto R(q_1,q_2)$ is an injection (when it is defined).

So because for a given $R$ , $r_2$ has only one possible value, there are at most two rational points.

Now to prove that $2$ is the answer, I will give an example.

Let $q_1=2$ and $q_2=5$ .

So $r_2$ has to be $-1$ and $r_1^2$ has to be $\frac{4q_2-q_1^2}{4}=4$ ie $r_1=\pm 2$ .

Verification:

$\begin{aligned} r_1^2+(r_2-\pi)^2 & = 4 + (-1-\pi)^2\\ & = 4 + 1 +2\pi +\pi^2\\ & =\pi^2 +2\pi + 5\\ & = R^2 \ \checkmark \end{aligned}$

The points $(2,-1)$ and $(-2,-1)$ are two distinct rational points on a circle center at $(0,\pi)$ .