Respuesta :

[tex]\bf \textit{equation of a circle}\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 \qquad center~~(\stackrel{}{{{ h}}},\stackrel{}{{{ k}}})\qquad \qquad radius=\stackrel{}{{{ r}}}\\\\ -------------------------------\\\\ (x-5)^2+(y+3)^2=16\implies [x-\stackrel{h}{5}]^2+[y-(\stackrel{k}{-3})]^2=\stackrel{r~~2}{4}[/tex]
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