so, since we know the coordinates for the diameter segment, let's see how long it is, recall that the radius is half of the diameter, thus
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -8 &,& -6~) % (c,d) &&(~ -4 &,& -14~) \end{array}\\\\\\ d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ d=√([-4-(-8)]^2+[-14-(-6)]^2) \\\\\\ d=√((-4+8)^2+(-14+6)^2)\implies d=√(4^2+(-8)^2) \\\\\\ d=√(80)\qquad \textit{therefore the radius is }\qquad r=\cfrac{√(80)}{2}]()
now, since the diameter goes through the center of the circle, the midpoint of it, will be the center of the circle then,
![\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -8 &,& -6~) % (c,d) &&(~ -4 &,& -14~) \end{array}~~ % coordinates of midpoint \left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-4-8}{2}~~,~~\cfrac{-14-6}{2} \right)\implies (-6~,~-10)]()
and since we know the coordinates for the center and the radius, let's plug those two in
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-6}{ h},\stackrel{-10}{ k})\qquad \qquad radius=\stackrel{(√(80))/(2)}{ r} \\\\\\\ [x-(-6)]^2+[y-(-10)]^2=\left( (√(80))/(2) \right)^2 \\\\\\ (x+6)^2+(y+10)^2=\cfrac{(√(80))^2}{2^2}\implies (x+6)^2+(y+10)^2=\cfrac{80}{4} \\\\\\ (x+6)^2+(y+10)^2=20](https://img.qammunity.org/2019/formulas/mathematics/college/32lbsc4s6m2ubk5ok50jpjupfi0kftmfum.png)