Respuesta :

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

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good job