Respuesta :
well, we know the endpoints of its diameter, so hmmm its center will be the midpoint of those endpoints.
[tex]~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-11}~,~\stackrel{y_1}{-9})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 1 -11}{2}~~~ ,~~~ \cfrac{ -3 -9}{2} \right)\implies \left( \cfrac{-10}{2}~~,~~\cfrac{-12}{2} \right)\implies \stackrel{center}{(-5~~,~~-6)}[/tex]
well, to get its radius, we can simply get the distance between both points and keeping in mind that the radius is half the diameter, we'll take half of that distance.
[tex]~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-11}~,~\stackrel{y_1}{-9})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{-3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[1 - (-11)]^2 + [-3 - (-9)]^2}\implies d=\sqrt{(1+11)^2+(-3+9)^2} \\\\\\ d=\sqrt{12^2 + 6^2}\implies d=\sqrt{180}\implies d=6\sqrt{5}~\hfill \underset{half~that}{\stackrel{radius}{3\sqrt{5}}}[/tex]
