<s>[INST] Simple algebra question concernig the equation of a circle for some reason, I can't do the math. I have worked out that $(x^2+y^2-1)^2+4y^2=r^2((x+1)^2 + y^2)^2$ should be the circle $$\left(x-\dfrac {1+r^2}{1-r^2}\right)^2 + y^2= \left(\dfrac{2r}{1-r^2}\right)^2,$$ using some prior knowledge, however I can't prove it. Can someone help me on my way? Many thanks in advance.
 [/INST] Consider the left-hand side as a quadratic equation in $y^2$:
$$ \begin{aligned}
(x^2+y^2-1)^2+4y^2 &= y^4 + 2y^2(x^2-1) + (x^2-1)^2 + 4 y^2 \\
  &= y^4 + 2y^2(x^2+1) + (x^2-1)^2 \\
 &= \left(y^2 + (x^2 + 1)\right)^2 + (x^2-1)^2 -  (x^2+1)^2 \\
 &= \left(y^2 + (x^2 + 1)\right)^2  - 4 x^2 \\
 &= \left(y^2 + (x + 1)^2\right)\left(y^2 + (x - 1)^2\right)
\end{aligned}$$
Then your equation becomes
$$
\left(y^2 + (x + 1)^2)\right)\left(y^2 + (x - 1)^2)\right)
 = r^2\left((x+1)^2 + y^2 \right)^2
$$
and it follows that 
$$
 y^2 + (x + 1)^2 = 0 
$$
which implies that $x=-1, y=0$, 
or the common factor can be cancelled:
$$
y^2 + (x - 1)^2 = r^2\left((x+1)^2 + y^2 \right) \\
\Longleftrightarrow
 (x^2+y^2+1)(1-r^2)-2x(1+r^2) = 0
$$
If $r=1$ then the solution is the line $x=0$ (i.e. the y-axis), otherwise 
the circle equation can now easily be derived.
</s>