if the endpoints of the diameter chord are P and Q, then the center must be in middle of that diameter, namely the midpoint of PQ.
[tex]\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
P(\stackrel{x_1}{-7}~,~\stackrel{y_1}{-4})\qquad
Q(\stackrel{x_2}{-1}~,~\stackrel{y_2}{10})
\qquad
% coordinates of midpoint
\left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right)
\\\\\\
\left( \cfrac{-1-7}{2}~~,~~\cfrac{10-4}{2} \right)\implies \left( \cfrac{-8}{2}~~,~~\cfrac{6}{2} \right)\implies (-4~,~3)[/tex]
