How do you convert the rectangular equation #x^2 + y^2 = 6x# to polar form? Precalculus Polar Equations of Conic Sections Writing Polar Equations for Conic Sections 1 Answer Alan P. Dec 6, 2015 #r=6cos(theta)# Explanation: #x^2+y^2 = r^2color(white)("XXXXXXXX")#[1] #x/r = cos(theta)# #rarr x=r*cos(theta)# #rarr 6x = 6r*cos(theta)color(white)("XXXX")#[2] Combining [1] and [2] #x^2+y^2=6x# #rarr r^2 = 6r*cos(theta)# #rarr r= 6cos(theta)# Answer link Related questions How do you identify conic sections? What is the meaning of conic section? What is the standard equation of a circle? What is the standard equation of a parabola? What is the standard equation of a hyperbola? Which conic section has the polar equation #r=1/(1-cosq)#? Which conic section has the polar equation #r=2/(3-cosq)#? Which conic section has the polar equation #r=a sintheta#? How do you find a polar equation for the circle with rectangular equation #x^2+y^2=25#? What are the polar coordinates of #(x-1)^2-(y+5)^2=-24#? See all questions in Writing Polar Equations for Conic Sections Impact of this question 11725 views around the world You can reuse this answer Creative Commons License