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Answer 1 · 2017-08-03T18:35:04

#(x^2-3x+y^2)^2=9(x^2+y^2)#.

#r=3(1+cos(theta))#

In general,

#x=rcos(theta)#, #y=rsin(theta)#.
Squaring and adding gives #r^2=x^2+y^2#.

You have
#r=3(1+cos(theta))#.

Then, #x=rcos(theta)#, giving #cos(theta)=x/r#.
#r=3(1+x/r)#,
#r^2=3(r+x)#,
#r^2-3x=3r#.

Squaring gives,
#(r^2-3x)^2=9r^2#.

Replacing #r^2# by #x^2+y^2# gives a final relation between #x# and #y#,
#(x^2-3x+y^2)^2=9(x^2+y^2)#.