Respuesta :

[tex]\bf \textit{equation of a circle}\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 \qquad center~~(\stackrel{3}{{{ h}}},\stackrel{6}{{{ k}}})\qquad \qquad radius=\stackrel{4}{{{ r}}} \\\\\\ (x-3)^2+(y-6)=4^2\implies (x-3)^2+(y-6)=16[/tex]

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