How do you convert #x^2+(y-4)^2=16# into polar form? Precalculus Polar Coordinates Converting Equations from Polar to Rectangular 1 Answer Bdub Sep 11, 2017 see below Explanation: Use the formulas #r^2=x^2+y^2, x= r cos theta, y= r sin theta# #x^2+(y-4)^2 = 16# -->FOIL #x^2+y^2-8y+16=16# #x^2+y^2-8y=16-16# #x^2+y^2-8y=0# #(x^2+y^2)-8y=0# #r^2-8rsin theta=0# 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 3503 views around the world You can reuse this answer Creative Commons License