well, we know the endpoints for the diameter, keeping in mind that the center of the circle will be in the halfway of it, thus its midpoint,
![\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -8 &,& 0~) % (c,d) &&(~ -12 &,& 2~) \end{array}\qquad % coordinates of midpoint \left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-12-8}{2}~~,~~\cfrac{2-0}{2} \right)\implies (-10~~,~~1)]()
so that's where the center is at.
keeping in mind that the radius is half the length of the diameter, what's the length of the diameter anyway?
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -8 &,& 0~) % (c,d) &&(~ -12 &,& 2~) \end{array}\\\\\\ d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ d=√([-12-(-8)]^2+[2-0]^2)\implies d=√((-12+8)^2+(2-0)^2) \\\\\\ d=√((-4)^2+2^2)\implies d=√(20)\qquad \qquad \qquad \qquad \stackrel{\textit{so the radius is then}}{\cfrac{√(20)}{2}}]()
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-10}{ h},\stackrel{1}{ k})\qquad \qquad radius=\stackrel{(√(20))/(2)}{ r}\\\\ -------------------------------\\\\\ [x-(-10)]^2+[y-1]^2=\left( (√(20))/(2) \right)^2\implies (x+10)^2+(y-1)^2=\cfrac{(√(20))^2}{2^2} \\\\\\ (x+10)^2+(y-1)^2=\cfrac{20}{4}\implies (x+10)^2+(y-1)^2=5](https://img.qammunity.org/2019/formulas/mathematics/college/omvmqtpoxxnctd6nrqzfbraasfuuywpl71.png)