User:Obradovic Goran

$$y=\frac{\sqrt{9-x^2}}{\sqrt{9-x^2}}-1; x=\frac{\sqrt{9-y^2}}{\sqrt{9-y^2}}-1$$ $$(x\pm2)^2+(y\pm2)^2=1.5^2, x\in(-3, 3)$$ So what if I'm a geek :)

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da:Bruger:Obradovic Goran sl:Uporabnik:Obradovic Goran sr:Корисник:Обрадовић Горан