How do you convert #-y+2=(3x+4)^2+(y-1)^2# into polar form? Trigonometry The Polar System Converting Between Systems 1 Answer Harish Chandra Rajpoot Jul 1, 2018 Given that #-y+2=(3x+4)^2+(y-1)^2# Substituting #x=r\cos\theta# & #y=r\sin\theta# #-r\sin\theta+2=(3r\cos\theta+4)^2+(r\sin\theta-1)^2# #-r\sin\theta+2=9r^2\cos^2\theta+16+24r\cos\theta+r^2\sin^2\theta+1-2r\sin\theta# #(8\cos^2\theta+1)r^2+(24\cos\theta-\sin\theta)r+15=0# Answer link Related questions How do you convert rectangular coordinates to polar coordinates? When is it easier to use the polar form of an equation or a rectangular form of an equation? How do you write #r = 4 \cos \theta # into rectangular form? What is the rectangular form of #r = 3 \csc \theta #? What is the polar form of # x^2 + y^2 = 2x#? How do you convert #r \sin^2 \theta =3 \cos \theta# into rectangular form? How do you convert from 300 degrees to radians? How do you convert the polar equation #10 sin(θ)# to the rectangular form? How do you convert the rectangular equation to polar form x=4? How do you find the cartesian graph of #r cos(θ) = 9#? See all questions in Converting Between Systems Impact of this question 1240 views around the world You can reuse this answer Creative Commons License