Question 950791
Assume that this point (a,b) is the center of a circle and the points lie on the circle. 
The circle has equation,
{{{x^2+fx+y^2+gy+h=0}}}
Input each point,
{{{4+2f+36+6g+h=0}}}
1.{{{2f+6g+h=-40}}}
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.
{{{9-3f+1+g+h=0}}}
2.{{{-3f+g+h=-10}}}
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.
{{{64+8f+36-6g+h=0}}}
3.{{{8f-6g+h=-100}}}
Now subtract eq. 1 from eq. 2 to eliminate {{{h}}},
{{{-3f+g+h-2f-6g-h=-10+40}}}
{{{-5f-5g=30}}}
4.{{{f+g=-6}}}
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.
{{{8f-6g+h-2f-6g-h=-100+40}}}
{{{6f-12g=-60}}}
5.{{{f-2g=-10}}}
Subtract eq. 4 from eq. 5,
{{{f-2g-f-g=-10-(-6)}}}
{{{-3g=-4}}}
{{{g=4/3}}}
Then,
{{{f+4/3=-6}}}
{{{f=-18/3-4/3}}}
{{{f=-22/3}}}
and finally,
{{{-3f+g+h=-10}}}
{{{22+4/3+h=-10}}}
{{{h=-30/3-4/3-66/3}}}
{{{h=-100/3}}}
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{{{x^2-(22/3)x+y^2+(4/3)y-100/3=0}}}
So now complete the square to find the center (h,k).
{{{x^2-(22/3)x+(11/3)^2+y^2+(4/3)y+(2/3)^2=100/3+(11/3)^2+(2/3)^2}}}
{{{(x-11/3)^2+(y+2/3)^2=300/9+121/9+4/9}}}
{{{(x-11/3)^2+(y+2/3)^2=425/9}}}
The center is ({{{11/3}}},{{{-2/3}}}).