How do you graph #r=1+2costheta#?

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Answer 1 · 2016-12-14T12:25:01

The graph of this limacon is obtained using the cartesian form
#(x^2+y^2)=sqrt(x^2+y^2)+2x#. Look at the dimple, at the pole r = 0..

Use the conversion formula #r(cos theta, sin theta) = (x, y)#, that

gives #cos theta =x/r and r=sqrt(x^2+y^2)#.

Hence #r=1+2costheta=>r^2=r+2rcostheta#

or #x^2+y^2=sqrt(x^2+y^2)+2x#

Then use the Socratic graphic facility.

The period for #r(theta)# is #2pi#. You can choose #theta in [0,

2pi]#.

graph{x^2+y^2-sqrt(x^2+y^2)-2x=0 [-5, 5, -2.5, 2.5]}