Respuesta:
[tex]5x^{2} +5y^{2} =25\\(x-a)^{2} +(y-b)^{2} =r^{2}\\(x-0)^{2}+(y-0)^{2} = (\sqrt{5})^{2} \\(a,b)=(0,0) r= \sqrt{5}[/tex]
Espero te sirva :V
Me das corona plis
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Respuesta:
[tex]5x^{2} +5y^{2} =25\\(x-a)^{2} +(y-b)^{2} =r^{2}\\(x-0)^{2}+(y-0)^{2} = (\sqrt{5})^{2} \\(a,b)=(0,0) r= \sqrt{5}[/tex]
Espero te sirva :V
Me das corona plis