What is the polar form of # x^2 + y^2 = 2x#?

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Answer 1 · 2014-11-12T20:34:09

#x^2+y^2=2x#, which looks like:

enter image source here

by plugging in #{(x=rcos theta),(y=rsin theta):}#,

#=> (rcos theta)^2+(r sin theta)^2=2rcos theta#

by multiplying out,

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

by factoring out #r^2# from the left-hand side,

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

by #cos^2theta+sin^2theta=1#,

#=> r^2=2rcos theta#

by dividing by #r#,

#=> r=2cos theta#, which looks like:

enter image source here

As you can see above, #x^2+y^2=2x# and #r=2cos theta# give us the same graphs.

I hope that this was helpful.