How do you graph #r=2+sintheta#?

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Answer 1 · 2016-10-10T21:54:59

#(x^2+y^2-y)^2-4(x^2+y^2)=0#

Using the pass equations

#{(x=rcostheta),(y=rsintheta):}#

we have

#r=2+y/r# or

#r^2=2r+y# or

#x^2+y^2-y=2sqrt(x^2+y^2)# squaring both sides

#(x^2+y^2-y)^2-4(x^2+y^2)=0#

Attached a plot

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