Respuesta :

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

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