Circumscribed Circle Calculator
Find the one circle through three points, or around a triangle from its three sides: circumradius, centre coordinates, areas and where the centre falls.
Circumscribed Circle Calculator
Background.
Three points that do not sit on one straight line always lie on exactly one circle. Finding it is a practical job as often as a textbook one: locating the centre of a bolt pattern from three measured holes, or recovering the radius of a curved wall from three surveyed points.
Enter the points as x and y coordinates and you get the circle's radius, its centre in the same coordinate system, the circle's area and circumference, the triangle's own area, and whether the centre lies inside, on or outside that triangle. If you only know the three side lengths, the shape is fixed but its position is not, so the triangle is placed with corner A at the origin and side c along the positive x-axis. Radius and areas are unaffected by that choice; the centre coordinates are measured in it.
Three points on a line lie on no circle, two identical points lie on infinitely many, and side lengths where one equals or exceeds the other two combined never close into a triangle, so none of these give a result.
What is circumscribed circle calculator?
The circumscribed circle, or circumcircle, of a triangle is the circle passing through all three corners. Its centre, the circumcentre, is the point equally distant from the three vertices, and that common distance is the circumradius R. Euclid constructs it in Elements Book IV, Proposition 5: bisect two sides, raise perpendiculars at their midpoints, and the perpendiculars meet at the centre. It works because every point on a segment's perpendicular bisector is the same distance from both ends of that segment.
How to use this calculator.
- Under "What do you have?" pick Three points for coordinates, or Three side lengths if you only measured the sides.
- For points, enter x and y for A, B and C in one consistent unit; negatives and decimals are fine.
- For sides, enter a, b and c, each named after the corner opposite it. Each must be shorter than the other two combined.
- Read the centre as where the compass point goes and the radius as how far to open it.
The formula.
With coordinates, the centre comes from solving the two perpendicular-bisector equations at once. Let d = 2[x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)] and sᵢ = xᵢ² + yᵢ². Then the centre is ux = [s₁(y₂ − y₃) + s₂(y₃ − y₁) + s₃(y₁ − y₂)] ÷ d and uy = [s₁(x₃ − x₂) + s₂(x₁ − x₃) + s₃(x₂ − x₁)] ÷ d, and R is the distance from (ux, uy) to any vertex. The quantity d is four times the triangle's signed area, so the triangle's area is |d| ÷ 4, and d = 0 is exactly the collinear case where no circle exists.
With side lengths, the triangle is first laid out as A = (0, 0), B = (c, 0) and C = (b cos A, b sin A), with cos A = (b² + c² − a²) ÷ 2bc from the law of cosines, and the same coordinate solution follows. The radius matches MathWorld's side-only formula R = abc ÷ √((a + b + c)(b + c − a)(c + a − b)(a + b − c)), which is abc ÷ 4K with the area K written out by Heron's formula.
The location line compares squared sides: with h the longest side and p, q the others, p² + q² − h² is positive for an acute triangle (centre inside), zero for a right triangle (centre on the hypotenuse's midpoint) and negative for an obtuse one (centre outside).
A worked example.
Take the points A = (0, 0), B = (4, 0) and C = (0, 3), the corners of a 3-4-5 right triangle. First, d = 2[0 × (0 − 3) + 4 × (3 − 0) + 0 × (0 − 0)] = 24. The squared distances from the origin are s₁ = 0, s₂ = 16 and s₃ = 9. The centre is then ux = (16 × 3) ÷ 24 = 2 and uy = (9 × 4) ÷ 24 = 1.5, so it sits at (2, 1.5). Its distance to A is √(2² + 1.5²) = √6.25 = 2.5, the circumradius. From R = 2.5 the circle's area is π × 6.25 = 19.635 and its circumference is 2π × 2.5 = 15.708. The triangle's area is 24 ÷ 4 = 6, so it covers about 31% of the disc. The location line reports a right triangle with the centre on the midpoint of the hypotenuse, and that is easy to confirm: the hypotenuse runs from (4, 0) to (0, 3), its midpoint is (2, 1.5) and half its length of 5 is 2.5.
Frequently asked questions.
How do I find the centre of an arc from three points measured along it?
Why does the centre sometimes land outside the triangle?
Why is the centre of a right triangle always on the hypotenuse?
Is the circumcentre the same as the centroid?
Can I find a circle through four or more points?
References& sources.
- [1]Wolfram Research, MathWorld — ‘Circumradius’ (live revision retrieved 2026-07-29). Equation (1), R = abc/√((a+b+c)(b+c−a)(c+a−b)(a+b−c)), and Equation (2), the semiperimeter form. Recorded honestly: the widely quoted R = abc/(4K) does not appear on that page in that arrangement — Equation (1) is the same identity with Heron's 4K written out — so this page cites what is actually printed. Equation (1) agrees with this calculator's coordinate-derived radius. Open access.
- [2]Euclid, Elements, Book IV, Proposition 5 — ‘To circumscribe a circle about a given triangle’, constructed by bisecting two sides and intersecting the perpendiculars raised at their midpoints. The accompanying guide names the circumcircle, circumcentre and circumradius, and states that the centre falls inside acute triangles, on the hypotenuse of right triangles, and outside obtuse triangles — the source of this page's location output. David E. Joyce web edition of the Heath translation, Clark University; confirmed on retrieval 2026-07-29.
- [3]Wolfram Research, MathWorld — ‘Heron's Formula’ (live revision retrieved 2026-07-29). Equation (2), Δ = √(s(s − a)(s − b)(s − c)). Used to cross-check the triangle area reported in side-lengths mode against the determinant used in coordinate mode, and to show that Equation (1) above is the familiar abc/4K. Open access.
- [4]Wolfram Research, MathWorld — ‘Polygon Area’ (live revision retrieved 2026-07-29). Equation (4), A = ½ Σ (xᵢyᵢ₊₁ − xᵢ₊₁yᵢ). Source for the triangle-area determinant, which is also the exact collinearity test this page uses to reject three points that lie on one line. Open access.
How this page was produced
- Published by
- Quanta Calculator
- Overseen by
- Daniel, Quanta Calculator Researcher
- Primary sources
- 4 cited below
- Method
- R = abc ⁄ √((a+b+c)(b+c−a)(c+a−b)(a+b−c)) , centre = perpendicular bisectors
- 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.
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