Question 1060451
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change to polar form: x^2+y^2= -2x
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In polar form  {{{x^2 + y^2}}} = {{{r^2}}}, {{{x = r*cos(alpha)}}}.

So, the original equation becomes

{{{r^2}}} = {{{-2rcos(alpha)}}},  or, canceling "r" in both sides,

r = {{{-2cos(alpha)}}}.


It is the equation in polar form.
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