Question 388559
The midpoint of (-6, 3) and (-4,-1) is (-5, 1).  The slope of the line passing through (-6, 3) and (-4,-1) is (3--1)/(-6--4) = 4/-2 = -2.  The equation of the line passing through the midpoint and perpendicular to the line through (-6, 3) and (-4,-1) is y-1 = (x--5)/2, or {{{y = (x+5)/2 + 1}}}.
The midpoint of (-2,5) and (-4,-1) is (-3, 2).  The slope of the line passing through (-2, 5) and (-4,-1) is (5--1)/(-2--4) = 6/2 = 3.  The equation of the line passing through the midpoint and perpendicular to the line through (-2, 5) and (-4,-1) is y-2 = -(x--3)/3, or {{{y = -(x+3)/3 + 2}}}.
The intersection of the two perpendicular bisectors is the center of the circle.
{{{(x+5)/2 + 1} = -(x+3)/3 + 2}}}.  Multiplying both sides by 6, we get {{{3x+15 + 6 = -2x-6+12}}}, or 3x+21 = -2x + 6, or 5x = -15, or x = -3.  To solve for y substitute into any one of the equations for the perpendicular bisector.  The radius is obtained by getting the distance of the center from any one of the given points.