Circumcircle of Triangle

What is the circumcenter of triangle $ABC$ with vertices

$$A=(1, 4), B=(-2, 3), C=(5, 2)?$$

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A circumcenter, by definition, is the center of the circle in which a triangle is inscribed, For this problem, let $O=(a, b)$ be the circumcenter of $\triangle ABC.$ Then since the distances to $O$ from the vertices are all equal, we have $$\lvert \overline{AO} \rvert=\lvert \overline{BO} \rvert=\lvert \overline{CO} \rvert.$$ From the first equality, we have $$\begin{aligned} \lvert \overline{AO} \rvert^2&=\lvert \overline{BO} \rvert^2\\ (a-1)^2+(b-4)^2&=(a+2)^2+(b-3)^2\\ -2a+1-8b+16&=4a+4-6b+9\\ 3a+b&=2. \qquad (1) \end{aligned}$$ Similarly, from the second equality, we have $$\begin{aligned} \lvert \overline{BO} \rvert^2&=\lvert \overline{CO} \rvert^2\\ (a+2)^2+(b-3)^2&=(a-5)^2+(b-2)^2\\ 4a+4-6b+9&=-10a+25-4b+4\\ 7a-b&=8. \qquad (2) \end{aligned}$$ Taking $(1)+(2)$ gives $a=1,$ which in turn gives $b=-1.$

Therefore, the circumcenter of triangle $ABC$ is $O=(1, -1).$ $_\square$