if the diameter is 6, then its radius is half that, or 3, thus,
[tex]\bf \textit{equation of a circle}\\\\
(x- h)^2+(y- k)^2= r^2
\qquad
center~~(\stackrel{2}{ h},\stackrel{-5}{ k})\qquad \qquad
radius=\stackrel{3}{ r}
\\\\\\\
[x-2]^2+[y-(-5)]^2=3^2\implies (x-2)^2+(y+5)^2=9[/tex]
