Question 1066044
Center is the midpoint of the diameter.
{{{x=(8+12)/2=10}}}
{{{y=(4+(-4))/2=0}}}
  

Radius SQUARED is  {{{cross((8-12)^2+(4-(-4))^2=16+64=80)}}} {{{(1/2)^2*sqrt((12-8)^2+(-4-4)^2)^2}}}

{{{(1/4)(sqrt(16+64))^2}}}

{{{(1/4)80}}}

{{{20}}}



EQUATION:  {{{(x-10)^2+y^2=20}}}