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

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Answer 1 · 2016-06-24T13:56:29

#r+2 cos(theta)= 0#

Substituting the pass equations

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

we have

#r^2cos(theta)^2+r^2sin(theta)^2+2r cos(theta)=0#

Simplifying

#r+2 cos(theta)= 0#

See figure attached

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