How do you convert #r = 1 + 2costheta# into rectangular form?

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Answer 1 · 2016-09-30T10:18:54

#(x^2+y^2-2x)^2=x^2+y^2#

With the pass equations

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

#r = 1+2x/r->r^2=r+2x# so

#r^2-2x = r->(r^2-2x)^2=r^2# or

#(x^2+y^2-2x)^2=x^2+y^2#

Attached a plot

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