[tex]\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{3}{ h},\stackrel{-3}{ k})\qquad \qquad radius=\stackrel{2\sqrt{6}}{ r} \\\\[-0.35em] ~\dotfill\\\\\ [x-3]^2~~ + ~~[y-(-3)]^2~~ = ~~(2\sqrt{6})^2\implies (x-3)^2+(y+3)^2=24[/tex]
