+23245 If you graph these, you can notice that these form a right triangle; therefore these three points are all found on one circle.
The endpoints of a diameter of this circle are (2,6) and (8,10).
The midpoint between these two points is (5,8).
This point is the center of the circle; the circumcenter.
+118609 Thanks Geno, that is a excellent method.
It always pays to do a quick sketch and then facts like this are more likely to be seen.
Here is the super long methods.
I'd have let the centre of the circle be (x,y)
Then found the distances and said they had to be equal.
\(r^2=(x-6)^2+(y-2)^2\qquad(1)\\ r^2=(x-8)^2+(y-6)^2\qquad (2)\\ r^2=(x-8)^2+(y-10)^2 \qquad (3)\\~\\ (3)-(2)\\ 0=(y-6)^2-(y-10)^2\\ 0=y^2-12y+36-(y^2-20y+100)\\ 0=y^2-12y+36-y^2+20y-100\\ 0=8y-64\\ 0=y-8\\ y=8\\~\\ sub\;\; in\\ r^2=(x-6)^2+(6)^2\qquad(1b)\\ r^2=(x-8)^2+(2)^2\qquad (2b)\\ subtract\\ 0=(x-6)^2+36-(x-8)^2-4\\ 0=(x^2-12x+36)+36-(x^2-16x+64)-4\\ 0=-12x+36+36+16x-64-4\\ 0=4x+4\\ x=-1 \)
\(r^2=(x-2)^2+(y-6)^2\qquad(1)\\ r^2=(x-8)^2+(y-6)^2\qquad (2)\\ r^2=(x-8)^2+(y-10)^2 \qquad (3)\\~\\ (3)-(2)\\ 0=(y-6)^2-(y-10)^2\\ 0=y^2-12y+36-(y^2-20y+100)\\ 0=y^2-12y+36-y^2+20y-100\\ 0=8y-64\\ 0=y-8\\ y=8\\~\\ sub\;\; in\\ r^2=(x-2)^2+(0)^2\qquad(1b)\\ r^2=(x-8)^2+(0)^2\qquad (2b)\\ subtract\\ 0=(x-2)^2-(x-8)^2\\ 0=(x^2-4x+4)-(x^2-16x+64)\\ 0=-4x+4+16x-64\\ 0=12x-60\\ x=5\\~\\ \text{So the centre is (5,8)} \)
Coding:
r^2=(x-2)^2+(y-6)^2\qquad(1)\\
r^2=(x-8)^2+(y-6)^2\qquad (2)\\
r^2=(x-8)^2+(y-10)^2 \qquad (3)\\~\\
(3)-(2)\\
0=(y-6)^2-(y-10)^2\\
0=y^2-12y+36-(y^2-20y+100)\\
0=y^2-12y+36-y^2+20y-100\\
0=8y-64\\
0=y-8\\
y=8\\~\\
sub\;\; in\\
r^2=(x-2)^2+(0)^2\qquad(1b)\\
r^2=(x-8)^2+(0)^2\qquad (2b)\\
subtract\\
0=(x-2)^2-(x-8)^2\\
0=(x^2-4x+4)-(x^2-16x+64)\\
0=-4x+4+16x-64\\
0=12x-60\\
x=5\\~\\
\text{So the centre is (5,8)}
