Question 581654
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 (-6,1) and (4,-5), we need to find the midpoint.



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



Since the x coordinate of midpoint is {{{-1}}}, this means that {{{h=-1}}}



Y-Coordinate of Midpoint = {{{(y[1]+y[2])/2 = (1+-5)/2=-4/2 = -2}}}



Since the y coordinate of midpoint is {{{-2}}}, this means that {{{k=-2}}}



So the center is the point (-1, -2)



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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=-1}}} and {{{k=-2}}}. Also, {{{x=-6}}} and {{{y=1}}}. Plug these values into the equation above and simplify to get:



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



{{{r^2=(-5)^2+(3)^2}}}



{{{r^2=25+9}}}



{{{r^2=34}}}



So because  {{{h=-1}}}, {{{k=-2}}}, and {{{r^2=34}}}, this means that the equation of the circle that passes through the points (-6,1) and (4,-5) (which are the endpoints of the diameter) is 



{{{(x+1)^2+(y+2)^2=34}}}.



So the answer is {{{(x+1)^2+(y+2)^2=34}}}