Question 551091
I'm assuming that these are the endpoints of the diameter. 



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 (3,5) and (1,2), we need to find the midpoint.



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



Since the x coordinate of midpoint is {{{2}}}, this means that {{{h=2}}}



Y-Coordinate of Midpoint = {{{(y[1]+y[2])/2 = (5+2)/2=7/2}}}



Since the y coordinate of midpoint is {{{7/2}}}, this means that {{{k=7/2}}}



So the center is the point (2, 7/2)



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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=2}}} and {{{k=7/2}}}. Also, {{{x=3}}} and {{{y=5}}}. Plug these values into the equation above and simplify to get:



{{{r^2=(3-2)^2+(5-7/2)^2}}}



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



{{{r^2=1+9/4}}}



{{{r^2=13/4}}}



So because  {{{h=2}}}, {{{k=7/2}}}, and {{{r^2=13/4}}}, this means that the equation of the circle that passes through the points (3,5) and (1,2) (which are the endpoints of the diameter) is 



{{{(x-2)^2+(y-7/2)^2=13/4}}}.