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

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Answer 1 · 2016-05-08T17:24:31

#r^2 = r cos theta#

To convert to polar form, substitute:

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

then simplify:

So:

#x^2+y^2 = x#

becomes:

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

The left hand side simplifies as follows:

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

#=r^2 cos^2 theta + r^2 sin^2 theta#

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

#=r^2#

So:

#r^2 = r cos theta#

You might be tempted to simplify this by dividing both sides by #r# to get:

#r = cos theta#

but that loses the solution #r = 0#, so it is probably best left in the form:

#r^2 = r cos theta#