Question 374780
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 (-2,4) and (4,2), we need to find the midpoint.



X-Coordinate of Midpoint = {{{(x[1]+x[2])/2 = (-2+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 = (4+2)/2=6/2 = 3}}}



Since the y coordinate of midpoint is {{{3}}}, this means that {{{k=3}}}



So the center is the point (1, 3)



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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=3}}}. Also, {{{x=-2}}} and {{{y=4}}}. Plug these values into the equation above and simplify to get:



{{{r^2=(-2-1)^2+(4-3)^2}}}



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



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



{{{r^2=10}}}



So because  {{{h=1}}}, {{{k=3}}}, and {{{r^2=10}}}, this means that the equation of the circle that passes through the points (-2,4) and (4,2) (which are the endpoints of the diameter) is 



{{{(x-1)^2+(y-3)^2=10}}}.



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Jim