How do you convert #x^2 +y^2-2x=0# to polar form?

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Answer 1 · 2016-05-21T17:33:04

#2rcos theta = r^2#

Substitute:

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

So we have:

#0 = x^2+y^2-2x#

#=(r cos theta)^2+(r sin theta)^2-2r cos theta#

#=r^2(cos^2 theta + sin^2 theta) - 2r cos theta#

#=r^2-2r cos theta#

Add #2rcos theta# to both ends and transpose to find:

#2rcos theta = r^2#

graph{x^2+y^2-2x=0 [-1.667, 3.333, -1.28, 1.22]}