Find rational numbers $(x,y)$ such that $ (x^2 + y^2 - 2x)^2 = x^2 + y^2$ The general limaçon has both a polar equation: $r = b + a \cos \theta  $ and an algebraic equation:
$$ (x^2 + y^2 - ax)^2 = b^2 (x^2 + y^2)$$
Can we find all the rational points on a curve like this?
I want to consider the case $a = 2b$ and $b = 1$:
$$ (x^2 + y^2 - 2x)^2 = x^2 + y^2$$
One solution is $(x,y) = (0,1)$ and another is $(x,y) = (1,0)$. How can I generate the other solutions over $\mathbb{Q}$ ?


An even easier route for generating rational points is to start from the parametric equations of the limaçon: $\left((1+2\cos\theta)\cos\theta,(1+2\cos\theta)\sin\theta\right)^\top$, and then perform the Weierstrass substitution $\cos\theta\mapsto\frac{1-u^2}{1+u^2},\;\sin\theta\mapsto\frac{2u}{1+u^2}$. Due to the nature of the substitution, you will not be able to obtain the point $(1,0)$, but all other rational points correspond to rational values of $u$.
