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Answer 1 · 2017-01-08T09:02:59

$(x^2+y^2)=3sqrt(x^2+y^2)-x$

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

where $r = sqrt(x^2+y^2)>=0$ .

After substitutions and rearrangement,

Seemingly a circle, the graph is for a limacon that has an invisible

dent, on the right.

$(x^2+y^2)=3sqrt(x^2+y^2)-x$ .

graph{x^2+y^2-3sqrt(x^2+y^2)+x=0 [-10, 10, -5, 5]}