${\displaystyle y={\frac {\sqrt {9-x^{2}}}{\sqrt {9-x^{2}}}}-1;x={\frac {\sqrt {9-y^{2}}}{\sqrt {9-y^{2}}}}-1}$
${\displaystyle (x\pm 2)^{2}+(y\pm 2)^{2}=1.5^{2},x\in (-3,3)}$
So what if I'm a geek :)

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