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 $\mathbb{Q}$ ?


Cases in which $x$ or $y$ are $0$ are easily described, so suppose that neither are $0$.  
Define $\lambda$ by $y=\lambda x$. Since $x,y\in \mathbb Q$, with neither of them equal to $0$, we see that $\lambda \in \mathbb Q$.
Your equation becomes $$\left( (1+\lambda^2)x^2-2x\right)^2=(1+\lambda^2)x^2$$
Divide through by $x^2$ to get $$\left( (1+\lambda^2)x-2\right)^2=(1+\lambda^2)\implies x=\frac {2\pm \sqrt {1+\lambda^2}}{1+\lambda^2}$$
We deduce that we are looking for $\lambda \in \mathbb Q$ such that $\sqrt {1+\lambda^2}\in \mathbb Q$. But taking any rational Pythagorean triple $(a,b,c)$ gives the solution $\lambda =\frac ba$.  Conversely, given $\lambda=\frac mn\in \mathbb Q$ with $\sqrt {1+\lambda^2}=\frac rs\in \mathbb Q$, we have $$1+\frac {m^2}{n^2}=\frac {r^2}{s^2}\implies (ns)^2+(ms)^2=(rn)^2$$ so any rational point on your curve arises from a Pythagorean triple.
For example: starting with the triple $(3,4,5)$ gives us the solution $(\frac {33}{25},\frac {44}{25})$
