How do you convert $r = 1 + 2 cos θ$ $r = 1 + 2costheta$ into rectangular form?

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Answer 1 · 2016-09-30T10:18:54

${(x^{2} + y^{2} - 2 x)}^{2} = x^{2} + y^{2}$ $(x^2+y^2-2x)^2=x^2+y^2$

With the pass equations

${(x=r cos theta),(y=r sin theta):}$ we have

$r = 1+2x/r->r^2=r+2x$ so

$r^2-2x = r->(r^2-2x)^2=r^2$ or

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

Attached a plot

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How do you convert r = 1 + 2costhetar=1+2cosθr = 1 + 2costheta into rectangular form?