We consider the set of points given by $P_s:=\left(1,\frac{2s}{s^2-1}\right)$ for all rational $s \neq \pm 1$. Their distances from the origin are rational by $\sqrt{1^2+\left(\frac{2s}{s^2-1}\right)^2} = \sqrt{\frac{s^4-2s^2+1+4s^2}{(s^2-1)^2}}=\frac{s^2+1}{s^2-1}$, and trivially their pairwise distances are rational. If we invert on the unit circle and $A,B$ are points with images $A^\prime , B^\prime$, we have $\overline{A'B'}=\frac{\overline{AB}}{\overline{A0} \cdot \overline{B0}}$. Especially all $P_s$ have pairwise rational distances again and get, together with the whole line $x=1$, mapped to the circle around $\left(\frac{1}{2},0\right)$ with radius $\frac 12$. Thus their images solve the problem.

Very nice solution,Zetax!!!!! there is an agly one: Lemma:There exist infinite sequence $\beta_i$ such that $cos \beta_i,sin \beta_i$ are both rational.then the distences between any two point of $P_i(cos2 \beta_i,sin 2\beta_i)$ are rational

Peter wrote: Does there exist a circle and an infinite set of points on it such that the distance between any two points of the set is rational? Lemma There are infinitely many rational numbers $ q$ such that $ \sqrt{1+q^{2}}$ is also rational. Proof of Lemma We need to show that the curve $ 1+x^{2}=y^{2}$ has infinitely many rational points. Indeed, for all nonzero rational numbers $ q$, the rational point $ \left(\frac{q^{2}-1}{2q},\frac{q^{2}+1}{2q}\right)$ lies on the curve. Now, we prove the following statement. There exist infinitely many concyclic rational points such that the distance between any two points of the set is rational. Proof We construct a set $ \mathbb{U}$ satisfying the conditions: \[ \mathbb{U}=\left\{\left(\frac{1-q^{2}}{1+q^{2}},\frac{2q}{1+q^{2}}\right)\;\vert\;\, q\in\mathbb{Q},\,\sqrt{1+q^{2}}\in\mathbb{Q}\right\}.\] One may use Lemma to see that $ \mathbb{U}$ is infinite. Also, it is easy to check that $ \left(\frac{1-q^{2}}{1+q^{2}},\frac{2q}{1+q^{2}}\right)\in\mathbb{U}$ is a rational point on the circle $ x^{2}+y^{2}=1$. It remains to prove that any two points in $ \mathbb{U}$ has rational length. Let $ \mathbf{P}\left(\frac{1-p^{2}}{1+p^{2}},\frac{2p}{1+p^{2}}\right)\in\mathbb{U}$ and $ \mathbf{Q}\left(\frac{1-q^{2}}{1+q^{2}},\frac{2q}{1+q^{2}}\right)\in\mathbb{U}$. It is not difficult to check that \[ \overline{\mathbf{P}\mathbf{Q}}=\frac{2\,\vert p-q\vert}{\sqrt{1+p^{2}}\sqrt{1+q^{2}}}.\] Since $ p$, $ q$, $ \sqrt{1+p^{2}}$ and $ \sqrt{1+q^{2}}$ are all rational, it is clear that $ \overline{\mathbf{P}\mathbf{Q}}$ is also rational. Note In the solution, we used the formula \[ \overline{\mathbf{P}\mathbf{Q}}=\sqrt{\left(\frac{1-p^{2}}{1+p^{2}}-\frac{1-q^{2}}{1+q^{2}}\right)^{2}+\left(\frac{2p}{1+p^{2}}-\frac{2q}{1+q^{2}}\right)^{2}}=\frac{2\,\vert p-q\vert}{\sqrt{1+p^{2}}\sqrt{1+q^{2}}}.\] Here, we offer an outline of a trigonometric proof. After setting $ p=\tan{\alpha}$ and $ q=\tan{\beta}$, where $ \alpha,\beta\in\mathbb{R}$, we obtain $ \mathbf{P}(\cos 2\alpha,\sin 2\alpha)$ and $ \mathbf{Q}(\cos 2\beta,\sin 2\beta)$. Since $ \overline{\mathbf{O}\mathbf{P}}=\overline{\mathbf{O}\mathbf{Q}}=1$ and since $ \angle\mathbf{P}\mathbf{O}\mathbf{Q}= 2\,\vert\alpha-\beta\vert$, we obtain $ \overline{\mathbf{P}\mathbf{Q}}=2\,\left\vert\sin\left(\alpha-\beta\right)\right\vert$. It now remains to observe that \[ \left\vert\sin\left(\alpha-\beta\right)\right\vert =\left\vert\cos\alpha\cos\beta\left(\frac{\sin\alpha}{\cos\alpha}-\frac{\sin\beta}{\cos\beta}\right)\right\vert =\frac{\vert\tan\alpha-\tan\beta\vert }{\sqrt{ 1+{\tan\alpha}^{2}}\sqrt{ 1+{\tan\beta}^{2}}}.\]