Circumcenter Of A Triangle Calculator
Find a triangle's circumcenter and circumradius from the coordinates of its three vertices: the one point equally distant from all three corners.
Circumcenter Of A Triangle Calculator
Background.
Any three points that do not lie on one straight line sit on exactly one circle. The centre of that circle is the triangle's circumcenter, the single point the same distance from all three corners, and that shared distance is the circumradius.
On paper, you find it by drawing the perpendicular bisectors of the sides. Every point on one bisector is equally far from the two ends of that side, so the point where two bisectors cross is equally far from all three vertices, and the third bisector is forced through the same spot. In coordinates the construction becomes algebra: set the squared distances from an unknown centre to each vertex equal, let the squared terms cancel in pairs, and solve the two linear equations that remain. The page does this with a closed-form determinant formula.
Where the centre lands says something about the triangle. In an acute triangle it lies inside; in a right triangle it sits exactly on the midpoint of the hypotenuse; in an obtuse triangle it falls outside, beyond the longest side. A circumcenter outside the triangle is therefore not a mistake.
The same problem turns up well outside the classroom. Three marks on the rim of a broken plate, or on a curved wall, are enough to recover the full circle and its centre. In computational geometry, the circumcenters of the triangles in a Delaunay triangulation are the corners of the matching Voronoi diagram.
If the three points are collinear, no circle passes through them, and the page refuses them instead of returning a meaningless centre. Nearly collinear points are accepted, but expect a very distant centre and an enormous radius, because the circle through them is almost a straight line.
What is circumcenter of a triangle calculator?
The circumcenter of a triangle is the centre of its circumscribed circle, the unique circle that passes through all three vertices. It is the intersection of the perpendicular bisectors of the sides, and its distance to each vertex is the circumradius, R. For a triangle with sides a, b and c and area K, R = abc ÷ 4K. It is one of the classical triangle centres alongside the centroid, incenter and orthocenter, and in an equilateral triangle all four are the same point.
How to use this calculator.
- Enter the x and y coordinates of vertex A, then B, then C. The order of the vertices does not change the centre.
- Use one unit for all six numbers, whether metres, millimetres or grid squares; the radius comes out in that same unit.
- When recovering a circle from a real object, mark three points spread as widely around the arc as you can. Points bunched close together let small measuring errors move the centre a long way.
- Read the centre's x and y coordinates and the radius. A centre that lies outside the triangle simply means one of its angles is obtuse.
The formula.
The centre (Ux, Uy) must satisfy (Ux − x1)² + (Uy − y1)² = (Ux − x2)² + (Uy − y2)² = (Ux − x3)² + (Uy − y3)². Expanding and subtracting these removes Ux² and Uy², leaving two linear equations whose solution is Ux = [S1(y2 − y3) + S2(y3 − y1) + S3(y1 − y2)] ÷ D and Uy = [S1(x3 − x2) + S2(x1 − x3) + S3(x2 − x1)] ÷ D, where each Sᵢ = xᵢ² + yᵢ² and D = 2[x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)]. D equals four times the triangle's signed area, so it is zero precisely when the points are collinear, which is the case the page rejects. The radius is the distance from the centre to vertex A, √[(Ux − x1)² + (Uy − y1)²]. With D in the denominator, a thin triangle with a small area pushes the centre a long way off.
A worked example.
Take A = (0, 0), B = (6, 0) and C = (0, 8), a right triangle with its right angle at A. First the denominator: D = 2[0 × (0 − 8) + 6 × (8 − 0) + 0 × (0 − 0)] = 2 × 48 = 96. The squared lengths from the origin are S1 = 0, S2 = 36 and S3 = 64. Then Ux = [0 + 36 × (8 − 0) + 64 × (0 − 0)] ÷ 96 = 288 ÷ 96 = 3, and Uy = [0 + 36 × (0 − 0) + 64 × (6 − 0)] ÷ 96 = 384 ÷ 96 = 4. The circumcenter is (3, 4), and its distance to A is √(3² + 4²) = 5, the circumradius. That point is the midpoint of the hypotenuse from (6, 0) to (0, 8), as it must be for a right triangle, and the radius is half the hypotenuse of 10. Move C to (1, 1), making the angle at C obtuse, and the centre drops outside the triangle to (3, −2), with a radius of √13 ≈ 3.606.
Frequently asked questions.
Why is the circumcenter outside my triangle?
How do I find the centre of a circle from three points on its edge?
Can three points on a straight line have a circumcenter?
How is the circumcenter different from the centroid and the incenter?
Is there a quick way to check the circumradius from the side lengths?
References& sources.
- [1]OpenStax, Rice University. Precalculus 2e, 2021. Trigonometry and analytic geometry chapters. Retrieved 2026-08-06.
- [2]OpenStax, Rice University. College Algebra 2e, 2021. Algebra and function chapters. Retrieved 2026-08-06.
- [3]NIST/SEMATECH. e-Handbook of Statistical Methods, 2012. Chapter 1.3 and Chapter 4. Retrieved 2026-08-06.
How this page was produced
- Published by
- Quanta Calculator
- Overseen by
- Daniel, Quanta Calculator Researcher
- Primary sources
- 3 cited below
- Method
- Solve equal squared distances from the center to three noncollinear vertices
- Published
- Last verified
Built with AI assistance and verified by automated tests against the cited sources — every worked example on this page is computed by the same code that runs the calculator. How we build and check calculators.
In this category
Embed
Quanta Pro
Paid features are coming later.
- All 1560 calculators remain free
- No billing is enabled