What is the polar graph of r=1-cosx?

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Answer 1 · 2017-02-03T11:50:36

See the graph and the explanation.

graph{x^2+y^2=sqrt(x^2+y^2)-x2 [-5, 5, -2.5, 2.5]}

Using usual polar #theta#, instead of x,

#r = 1-cos theta>=0 and <=2#,

represents the curve that is befittingly named cardioid.

It has a dimple, at the pole r = 0.

For the Cartesian frame, the equation is

#x^2+y^2= r^2=rxxr=r(1-costheta)=r-rcostheta=sqrt(x^2+y^2)-x#,

giving

#x^2+y^2=sqrt(x^2+y^2)-x#, This is used for the readily available

Socratic utility.

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