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

Question

23703 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-31T09:54:21

two versions of the same cardioid

$(r+b-2acos(theta))(r-b-2acos(theta))=0$

Taking

$(x^2 + y^2 - 2 a x)^2 - b^2 (x^2 + y^2)=(x^2 + y^2 - 2 a x + b sqrt[x^2 + y^2])(x^2 + y^2 - 2 a x - b sqrt[x^2 + y^2]) = 0$

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

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