wait just a second, what's the radius of the circle anyway?
well, the radius is the segment going from the center of the circle to the circle itself, well, low and behold, if we just get the distance from those two points, that gives us the radius.
[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ -1 &,& -3~)
% (c,d)
&&(~ -7 &,& -5~)
\end{array}\\\\\\
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
r=\sqrt{[-7-(-1)]^2+[-5-(-3)]^2}\implies r=\sqrt{(-7+1)^2+(-5+3)^2}
\\\\\\
r=\sqrt{(-6)^2+(-2)^2}\implies r=\sqrt{40}\\\\
-------------------------------[/tex]
[tex]\bf \textit{equation of a circle}\\\\
(x- h)^2+(y- k)^2= r^2
\qquad
center~~(\stackrel{-1}{ h},\stackrel{-3}{ k})\qquad \qquad
radius=\stackrel{\sqrt{40}}{ r}
\\\\\\\
[x-(-1)]^2+[y-(-3)]^2=(\sqrt{40})^2\implies (x+1)^2+(y+3)^2=40[/tex]
