
The circumcenter of a triangle is the the point where the three perpendicular bisectors of its sides intersect.

The three perpendicular bisectors have to intersect in a single point because...
First, observe that any three noncollinear points determine a circle, called the circumcircle. Next, observe that each of the sides of the triangle is a chord of the circumcircle. Finally, observe that the perpendicular bisector of a chord is a diameter of the circle, and all diameters of a circle pass through the circle's center.
Alternatively, Carnot's Theorem can be used to prove this.
( a^{2}(b^{2}+c^{2}a^{2}), b^{2}(c^{2}+a^{2}b^{2}), c^{2}(a^{2}+b^{2}c^{2}) )
You can think of the barycentric coordinates of the "weights" of the vertices, so the weighted average of the vertices is the circumcenter. You can calculate the circumcenter from the barycentric coordinates by multiplying each vertex (vector) by the corresponding barycentric coordinate, adding the results, and then dividing by the sum of the barycentric coordinates.
Choosing vertices If the vertices are A(a,b), B(c,d), C(e,f), I will replace a^{2} by ((ce)^{2}+(df)^{2}), b^{2} by ((ae)^{2}+(bf)^{2}), and c^{2} by ((ac)^{2}+(bd)^{2}), and then the x coordinate of the circumcenter is given by
(a * ((ce)^{2}+(df)^{2})*(((ae)^{2}+(bf)^{2})+((ac)^{2}+(bd)^{2})((ce)^{2}+(df)^{2}))+
c * ((ae)^{2}+(bf)^{2})*(((ac)^{2}+(bd)^{2})+((ce)^{2}+(df)^{2})((ae)^{2}+(bf)^{2}))+
e * ((ac)^{2}+(bd)^{2})*(((ce)^{2}+(df)^{2})+((ae)^{2}+(bf)^{2})((ac)^{2}+(bd)^{2})) ) /
( ((ce)^{2}+(df)^{2})*(((ae)^{2}+(bf)^{2})+((ac)^{2}+(bd)^{2})((ce)^{2}+(df)^{2}))+
((ae)^{2}+(bf)^{2})*(((ac)^{2}+(bd)^{2})+((ce)^{2}+(df)^{2})((ae)^{2}+(bf)^{2}))+
((ac)^{2}+(bd)^{2})*(((ce)^{2}+(df)^{2})+((ae)^{2}+(bf)^{2})((ac)^{2}+(bd)^{2})) )which, I verified, simplifies to the value of h, given below, and I'm sure k is derived similarly.
The other (simpler) way to do this is to find the equation of a circle that passes through three points. If the vertices are A(a,b), B(c,d), C(e,f) then the circumcenter is given by (h,k), where h and k are
h = (1/2)((a�+b�)(fd) + (c�+d�)(bf) + (e�+f�)(db)) / (a(fd)+c(bf)+e(db))
k = (1/2)((a�+b�)(ec) + (c�+d�)(ae) + (e�+f�)(ca)) / (b(ec)+d(ae)+f(ca))
The radius of a circle circumscribed around a triangle is R = abc/(4K), where K is the area of the triangle.
Proof: Given a chord of a circle, the measure of the central angle subtended by that chord is twice the measure of an inscribed angle that the same chord subtends (proof). So the measure of angle AOB is twice the measure of ACB.
AOB is an isosceles triangle, so its altitude MO bisects angle AOB. So angle MOB is equal to angle ACB.
From right triangle MOB, c/2 = R sin(MOB), so sin(MOB)=c/(2R)
The area of triangle ACB, K, is given by
K = (1/2)ab sin(C)
K = (1/2)ab sin(MOB)
K = (1/2)abc/(2R)
K = abc/(4R)
R = abc/(4K)
Interesting fact: the orthocenter and the circumcenter (that is, the center of the circumscribed circle) are isogonal conjugates of one another.
Other triangle centers: Circumcenter, Incenter, Orthocenter, Centroid. The Orthocenter and Circumcenter of a triangle are isogonal conjugates, and the Incenter is its own isogonal conjugate.
Summary of geometrical theorems summarizes the proofs of concurrency of the lines that determine these centers, as well as many other proofs in geometry. In particular Carnot's Theorem can be used to show
Barycentric Coordinates, which provide a way of calculating these triangle centers (see each of the triangle center pages for the barycentric coordinates of that center).
Inscribed Angle Property  that all angles that are inscribed in a circle that are subtended by a given chord have equal measure, and that measure is half the central angle subtended by the same chord.
Law of Sines  Given triangle ABC with opposite sides a, b, and c, a/(sin A) = b/(sin B) = c/(sin C) = the diameter of the circumscribed circle.
Equation of a Circle given Three Points  if the three points are the vertices of a triangle, then the equation in question is that of the circumscribed circle.
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