Question #baaae

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Answer 1 · 2017-08-03T18:35:04

$(x^2-3x+y^2)^2=9(x^2+y^2)$ .

$r=3(1+cos(theta))$

In general,

$x=rcos(theta)$ , $y=rsin(theta)$ .
Squaring and adding gives $r^2=x^2+y^2$ .

You have
$r=3(1+cos(theta))$ .

Then, $x=rcos(theta)$ , giving $cos(theta)=x/r$ .
$r=3(1+x/r)$ ,
$r^2=3(r+x)$ ,
$r^2-3x=3r$ .

Squaring gives,
$(r^2-3x)^2=9r^2$ .

Replacing $r^2$ by $x^2+y^2$ gives a final relation between $x$ and $y$ ,
$(x^2-3x+y^2)^2=9(x^2+y^2)$ .