Check the picture below.
since we know the endpoints of the diameter, well, its midpoint will have to be the center of the circle, and half the length of the diameter is well, the radius.
[tex]\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\qquad B(\stackrel{x_2}{7}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{7+3}{2}~~,~~\cfrac{3+1}{2} \right)\implies \left(\cfrac{10}{2}~~,~~\cfrac{4}{2} \right)\implies (5~~,~~2) \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\qquad B(\stackrel{x_2}{7}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AB=\sqrt{(7-3)^2+(3-1)^2}\implies AB=\sqrt{4^2+2^2} \\\\\\ AB=\sqrt{20}~\hspace{10em}\stackrel{\textit{radius is half that}}{\cfrac{\sqrt{20}}{2}} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{5}{ h},\stackrel{2}{ k})\qquad \qquad radius=\stackrel{\frac{\sqrt{20}}{2}}{ r} \\\\\\ (x-5)^2+(y-2)^2=\left( \frac{\sqrt{20}}{2} \right)^2\implies (x-5)^2+(y-2)^2=\cfrac{20}{4} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (x-5)^2+(y-2)^2=5~\hfill[/tex]
