so hmm check the picture below, that's about the circle and the endpoints, but notice, the endpoints make up a segment, namely the diameter of the circle, well.... let's see how long that is, because, the radius is half the diameter
[tex]\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 7}}\quad ,&{{ 3}})\quad
% (c,d)
&({{ 7}}\quad ,&{{ -5}})
\end{array}\qquad
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
d=\sqrt{(7-7)^2+(-5-3)^2}\implies d=\sqrt{0+(-8)^2}\implies d=8
\\\\\\
\textit{the radius is half that, so is }\boxed{r=4}[/tex]
now.. hmmm notice, the midpoint of the diameter, is the center of the circle, let's check that one out
[tex]\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 7}}\quad ,&{{ 3}})\quad
% (c,d)
&({{ 7}}\quad ,&{{ -5}})
\end{array}\qquad
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left(\cfrac{{{ 7+7}}}{2}\quad ,\quad \cfrac{{{ -5}} + {{ 3}}}{2} \right)\implies \left( \cfrac{14}{2}\ ,\ \cfrac{-2}{2} \right)\implies \boxed{(7,-1)}[/tex]
now.. that we know what the center is, and what the radius is, well
[tex]\bf \textit{equation of a circle}\\\\
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad
\begin{array}{lllll}
center\ (&{{ h}},&{{ k\quad }})\qquad
radius=&{{ r}}\\
&7&-1&4
\end{array}[/tex]
