Taking
(x^2 + y^2 - 2 a x)^2 - b^2 (x^2 + y^2)=(x^2 + y^2 - 2 a x + b sqrt[x^2 + y^2])(x^2 + y^2 - 2 a x - b sqrt[x^2 + y^2]) = 0(x2+y2−2ax)2−b2(x2+y2)=(x2+y2−2ax+b√x2+y2)(x2+y2−2ax−b√x2+y2)=0
and then separately, after substituting
{
(x = r cos(theta)),
(y=r sin(theta))
:}
x^2 + y^2 - 2 a x + b sqrt[x^2 + y^2] = r+b-2a cos(theta)=0
x^2 + y^2 - 2 a x - b sqrt[x^2 + y^2] = r-b-2a cos(theta)=0
So, in polar coordinates reduces to
(r+b-2acos(theta))(r-b-2acos(theta))=0 wich are two versions of the same cardioid.
Attached a plot showing in red
r = -b + 2 a cos(theta), pi/3 < theta < pi/3
and in blue
r=b+2acos(theta), -2pi/3 < theta < 2pi/3
for ·a = b = 1
![enter image source here]()
