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

1 Answer
Jul 31, 2016

Answer:

two versions of the same cardioid

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

Explanation:

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