# What is the graph of the Cartesian equation (x^2 + y^2 - 2ax)^2 = b^2(x^2 + y^2)?

Jul 31, 2016

two versions of the same cardioid

$\left(r + b - 2 a \cos \left(\theta\right)\right) \left(r - b - 2 a \cos \left(\theta\right)\right) = 0$

#### Explanation:

Taking

${\left({x}^{2} + {y}^{2} - 2 a x\right)}^{2} - {b}^{2} \left({x}^{2} + {y}^{2}\right) = \left({x}^{2} + {y}^{2} - 2 a x + b \sqrt{{x}^{2} + {y}^{2}}\right) \left({x}^{2} + {y}^{2} - 2 a x - b \sqrt{{x}^{2} + {y}^{2}}\right) = 0$

and then separately, after substituting

{ (x = r cos(theta)), (y=r sin(theta)) :}

${x}^{2} + {y}^{2} - 2 a x + b \sqrt{{x}^{2} + {y}^{2}} = r + b - 2 a \cos \left(\theta\right) = 0$
${x}^{2} + {y}^{2} - 2 a x - b \sqrt{{x}^{2} + {y}^{2}} = r - b - 2 a \cos \left(\theta\right) = 0$

So, in polar coordinates reduces to

$\left(r + b - 2 a \cos \left(\theta\right)\right) \left(r - b - 2 a \cos \left(\theta\right)\right) = 0$ wich are two versions of the same cardioid.

Attached a plot showing in red
$r = - b + 2 a \cos \left(\theta\right) , \frac{\pi}{3} < \theta < \frac{\pi}{3}$
and in blue
$r = b + 2 a \cos \left(\theta\right) , - 2 \frac{\pi}{3} < \theta < 2 \frac{\pi}{3}$

for ·$a = b = 1$ 