Question 472993
Recall that the general equation of a circle is {{{(x-h)^2+(y-k)^2=r^2}}}.



So we need the center (h,k) and the radius squared {{{r^2}}}.



First, let's find the center (h,k).



Since the center is the midpoint of the line segment with endpoints (-6,-2) and (-4,4), we need to find the midpoint.



X-Coordinate of Midpoint = {{{(x[1]+x[2])/2 = (-6+-4)/2=-10/2 = -5}}}



Since the x coordinate of midpoint is {{{-5}}}, this means that {{{h=-5}}}



Y-Coordinate of Midpoint = {{{(y[1]+y[2])/2 = (-2+4)/2=2/2 = 1}}}



Since the y coordinate of midpoint is {{{1}}}, this means that {{{k=1}}}



So the center is the point (-5, 1)



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Now let's find the radius squared



Use the formula {{{r^2=(x-h)^2+(y-k)^2}}}, where (h,k) is the center and (x,y) is an arbitrary point on the circle.



In this case, {{{h=-5}}} and {{{k=1}}}. Also, {{{x=-6}}} and {{{y=-2}}} (drawn from one of the given points -- namely the first one). Plug these values into the equation above and simplify to get:



{{{r^2=(-6--5)^2+(-2-1)^2}}}



{{{r^2=(-6+5)^2+(-2-1)^2}}}



{{{r^2=(-1)^2+(-3)^2}}}



{{{r^2=1+9}}}



{{{r^2=10}}}



So because  {{{h=-5}}}, {{{k=1}}}, and {{{r^2=10}}}, this means that the equation of the circle that passes through the points (-6,-2) and (-4,4) (which are the endpoints of the diameter) is 



{{{(x+5)^2+(y-1)^2=10}}}.



So the answer is {{{(x+5)^2+(y-1)^2=10}}}