Circumscribed Circle Calculator

Background.

The Quanta circumscribed circle calculator finds the one circle that passes through three given points — equivalently, the circle drawn around a triangle so that all three corners sit on it. Give it three coordinate pairs, or just the three side lengths, and it returns the circumradius, the circumcentre, the circle's area and circumference, the triangle's own area, and a plain statement of whether that centre falls inside the triangle, on its edge, or outside it.

With the defaults — the points (0, 0), (4, 0) and (0, 3), the familiar 3-4-5 right triangle — the circumradius comes back as exactly 2.5 and the centre as exactly (2, 1.5). Neither is a rounded approximation. The triangle has a right angle at the origin, and Euclid's Book III, Proposition 31 says the angle in a semicircle is right, so the hypotenuse has to be a diameter: the centre is its midpoint and the radius is half its length, 5 ÷ 2. The circle's area is 19.6349540849 against the triangle's 6, so the triangle fills under a third of the disc.

Two things this page is careful about. First, the side-lengths mode has to put the triangle somewhere before it can report a centre, because three lengths fix a shape but not a position. It places vertex A at the origin with side c running along the positive x-axis, and the centre coordinates are relative to that placement — the radius and the two areas are not, and mean exactly what they say wherever the triangle sits. Second, the circumcentre is not the centroid. For this same default triangle the centroid, the balance point, is at (1.3333333333, 1), while the circumcentre is at (2, 1.5). They answer different questions, and the centroid calculator is linked for the other one.

What is circumscribed circle calculator?

The circumscribed circle, or circumcircle, of a triangle is the circle that passes through all three of its vertices. Euclid gives the construction at Elements Book IV, Proposition 5 — "to circumscribe a circle about a given triangle" — by bisecting two sides, erecting perpendiculars at those midpoints, and taking their intersection as the centre. Every point on the perpendicular bisector of a segment is equidistant from that segment's ends, so the intersection of two such bisectors is equidistant from all three vertices, and a circle drawn from there through one vertex necessarily passes through the other two. That point is the circumcentre and the common distance is the circumradius.

Any three points that do not lie on one straight line determine exactly one circle. Three collinear points determine none — or, depending on how you like to say it, a circle of infinite radius — which is why this calculator rejects them.

From the side lengths alone, MathWorld's Circumradius entry gives the closed form at Equation (1): R = abc ÷ √((a + b + c)(b + c − a)(c + a − b)(a + b − c)). The denominator is Heron's four-times-the-area written out in full, so this is the familiar R = abc ÷ 4K in the form it is actually printed. It is also how this page checks itself: the calculator reaches the answer by solving perpendicular bisectors in coordinates, and its test suite verifies the result against Equation (1) on every run, which is a genuinely separate derivation rather than a second reference for the same one.

The circumcentre moves as the triangle changes shape. Euclid IV.5's accompanying guide states the pattern: the centre falls inside an acute triangle, on the hypotenuse of a right triangle, and outside an obtuse one. This page reports which of the three you have, decided from the sign of p² + q² − h² where h is the longest side — the same quantity that decides the largest angle, computed from squared lengths so that no square root can blur the boundary.

How to use this calculator.

1. 01Choose whether you have coordinates or side lengths. Coordinates give a circumcentre in your own frame of reference; side lengths give the correct radius but a centre relative to a standard placement.
2. 02In coordinate mode, enter the three points. Negatives and decimals are fine. The three must be distinct and must not lie on one straight line.
3. 03In side-lengths mode, enter a, b and c. Each must be shorter than the sum of the other two — otherwise the lengths do not close into a triangle and no circle exists.
4. 04Read the circumradius at the top. That is the compass setting for drawing a circle through all three corners.
5. 05Read the circumcentre below it — that is where the compass point goes. In side-lengths mode, remember it is measured from vertex A at the origin with side c along the x-axis.
6. 06Check the last line to see where that centre falls. If it says obtuse, expect the compass point to land outside the triangle, which surprises people the first time.
7. 07If you wanted the balance point rather than the circle's centre, use the centroid calculator instead — for the default triangle those are (2, 1.5) and (1.3333333333, 1) respectively.

The formula.

In coordinate mode the calculator solves the two perpendicular-bisector equations directly. Writing sᵢ for xᵢ² + yᵢ² and d for 2[x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)], 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 the radius is the distance from that point to any one vertex. The quantity d is four times the signed area of the triangle, so d = 0 is exactly the collinear case — the calculator tests for it and refuses rather than dividing by zero.

In side-lengths mode the triangle is placed first: vertex A at the origin, vertex B at (c, 0), and vertex C at (b·cos A, b·sin A) with cos A = (b² + c² − a²) ÷ 2bc from the law of cosines. The identical coordinate code then runs. The radius and both areas are independent of where the triangle was put; the two centre coordinates are not, which is why the page says so beside the mode selector rather than leaving it to be discovered.

Working the default all the way through, with points (0, 0), (4, 0) and (0, 3): 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. So ux = [0 + 16 × 3 + 0] ÷ 24 = 48 ÷ 24 = 2 and uy = [0 + 0 + 9 × 4] ÷ 24 = 36 ÷ 24 = 1.5, both exact. The radius is the distance from (2, 1.5) to (0, 0), which is √(4 + 2.25) = √6.25 = 2.5, again exact. The circle's area is π × 6.25 = 19.6349540849 and its circumference is 2π × 2.5 = 15.7079632679. The triangle's area is 24 ÷ 4 = 6.

Two independent checks confirm those numbers. MathWorld's Equation (1) from the side lengths gives R = (5 × 3 × 4) ÷ √(12 × 2 × 6 × 4) = 60 ÷ √576 = 60 ÷ 24 = 2.5. And Euclid's Book III, Proposition 31 — the angle in a semicircle is right — means a right triangle's hypotenuse must be a diameter, so the centre is the midpoint of the segment from (4, 0) to (0, 3), namely (2, 1.5), with R equal to half of 5. That construction fixes both coordinates and the radius without using any formula at all.

The last output is a classification, and its scalar is stated explicitly: δ = p² + q² − h², where h is the longest side and p and q are the other two. Positive means acute and the centre lies inside; exactly zero means right and the centre lands on the midpoint of the hypotenuse; negative means obtuse and the centre falls outside. δ is built from squared side lengths and never from square roots, so for whole-number input it is exact — the default triangle gives 16 + 9 − 25 = 0 with no floating-point slack — and it classifies the unrounded value, so nothing that is displayed can contradict it.

Rounding stage: all arithmetic runs at 40 significant digits with π computed as arccos(−1), and rounding happens once, at the return boundary, to 10 decimal places. The circumcentre is the pivot: the radius is measured from the unrounded centre, and the circle's area and circumference are built from the unrounded radius. Rounding the centre first would push error into the ninth decimal place of the circle area, which is quadratic in R.

A worked example.

Frequently asked questions.

References& sources.

1. [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) is asserted against this calculator's coordinate-derived radius in circumscribed-circle.test.ts. Open access.
2. [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. [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. [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.