How this instrument works
The circumcenter is the one point in a triangle's plane sitting the same distance from all three vertices — pin a string to it and swing it around, and the string traces a circle passing exactly through the three corners. It's found by intersecting the perpendicular bisectors of the sides: the perpendicular bisector of any segment is precisely the set of points equidistant from that segment's two ends, so where two of the three bisectors cross must already be equidistant from all three vertices, and the third bisector passes through that same point automatically.
The formula above is just that intersection solved algebraically for two lines at once, with a shared denominator D equal to twice the triangle's signed area. When D is small the triangle is thin and needle-like; when D hits zero the three points have collapsed onto a single line, and no circle — hence no circumcenter — passes through them. That one equation quietly encodes the geometric fact that three collinear points never pin down a unique circle.
Where the circumcenter lands depends on the triangle's shape, and this is the detail that separates it from the centroid, the plain average of the three vertices, which always sits inside the shape no matter how it's stretched. In an acute triangle the circumcenter sits inside too; in an obtuse triangle it moves outside, past the longest side; and in a right triangle it lands exactly on the hypotenuse, at its midpoint — a fact worth confirming with the worked example below rather than taking on faith.
- Enter the first corner's coordinates in Vertex A: x and Vertex A: y.
- Enter the second corner in Vertex B: x and Vertex B: y, and the third in Vertex C: x and Vertex C: y — any labeling order works, as long as the three points aren't collinear.
- Read the result in Circumcenter: x and Circumcenter: y, the coordinates of the point equidistant from all three vertices.
- Change any of the six vertex fields to see both circumcenter outputs update, and watch how the point crosses outside the triangle once one angle passes 90°.
Worked example — the 3-4-5 right triangle
Take the right triangle with vertices A(0, 0), B(4, 0), and C(0, 3) — legs of length 4 and 3 along the axes, hypotenuse of length 5 running from B to C. Plugging into the formula: D = 2[0(0−3) + 4(3−0) + 0(0−0)] = 2(12) = 24. The cx numerator works out to (0)(−3) + (16)(3) + (9)(0) = 48, so cx = 48 ⁄ 24 = 2.0. The cy numerator works out to (0)(−4) + (16)(0) + (9)(4) = 36, so cy = 36 ⁄ 24 = 1.5.
The circumcenter (2.0, 1.5) lands exactly on segment BC, the hypotenuse, at its midpoint — (4+0) ⁄ 2 = 2 and (0+3) ⁄ 2 = 1.5 confirm it directly. That isn't a quirk of this particular triangle: for every right triangle, the hypotenuse is a diameter of the circumscribed circle, so its midpoint is always the circumcenter, a consequence of the rule that an angle inscribed in a semicircle is always a right angle.
Questions
What is the circumcenter of a triangle?
It's the point equidistant from all three vertices — equivalently, the center of the unique circle, called the circumcircle, that passes through all three corners. It's found where the perpendicular bisectors of the three sides cross, since a perpendicular bisector is exactly the set of points equidistant from a segment's two ends.
How is the circumcenter different from the centroid?
The centroid is the plain average of the three vertices and always sits inside the triangle. The circumcenter instead satisfies equal distance to the vertices, and depending on the angles it can sit inside, on an edge, or outside the triangle entirely. The two points coincide only when the triangle is equilateral.
Why can the circumcenter fall outside the triangle?
Because equidistant from the vertices is a different condition from inside the shape. In an obtuse triangle, the angle wider than 90 degrees pushes the equidistant point past the side that sits opposite it. In a right triangle the point lands exactly on the hypotenuse; only in an acute triangle does it stay inside.
What happens if the three points are collinear?
The formula divides by D, twice the triangle's signed area, and collinear points make D equal zero, so the calculation is undefined — no circle passes through three points sitting on one line. Enter nearly-collinear vertices and watch the circumcenter shoot far from the triangle as D approaches zero; that's the formula reflecting a genuine geometric limit, not an error.
How do the perpendicular bisectors prove the formula works?
Each perpendicular bisector is a line of points equidistant from one pair of vertices, so intersecting the bisectors of sides AB and BC gives a point equidistant from A, B, and C at once. Because that point already satisfies both conditions, it automatically lies on the third bisector too — which is why solving just two of the three bisector equations is enough.
Does the order I enter the three vertices in matter?
No. Relabeling which corner is A, B, or C describes the same triangle and returns the same circumcenter, since the formula depends only on the set of three points, not the labels attached to them. Entering the corners clockwise or counter-clockwise makes no difference to cx or cy.