Question 406017
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Hi
Midpoint ( {{{(x[1] + x[2])/2}}}, {{{(y[1] + y[2])/2}}})
(-4,6) to 
(8,-2)
Midpoint is (2,2) r = {{{sqrt(6^2 + 4^2) = sqrt(52)}}}
Standard Form of an Equation of a Circle is {{{(x-h)^2 + (y-k)^2 = r^2}}} 
where Pt(h,k) is the center and r is the radius
Circle: (x-2)^2 + (y-2)^2 = 52
 {{{drawing(300,300, -10,10,-10,10, grid(1),
circle(-4, 6,0.3),
circle(8, -2,0.3),
circle(2, 2,0.3),
circle(2, 2,7.21),
line(-4,6,8,-2),
graph( 300, 300, -10,10,-10,10,0))}}}