How do you convert #r = 1 + 2costheta# into rectangular form? Precalculus Polar Coordinates Converting Equations from Polar to Rectangular 1 Answer Cesareo R. Sep 30, 2016 #(x^2+y^2-2x)^2=x^2+y^2# Explanation: 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 Answer link Related questions What is the polar equation of a horizontal line? What is the polar equation for #x^2+y^2=9#? How do I graph a polar equation? How do I find the polar equation for #y = 5#? What is a polar equation? How do I find the polar equation for #x^2+y^2=7y#? How do I convert the polar equation #r=10# to its Cartesian equivalent? How do I convert the polar equation #r=10 sin theta# to its Cartesian equivalent? How do you convert polar equations to rectangular equations? How do you convert #r=6cosθ# into a cartesian equation? See all questions in Converting Equations from Polar to Rectangular Impact of this question 9204 views around the world You can reuse this answer Creative Commons License