About the Circumcircle & Incircle Calculator
Every triangle has exactly one circumscribed circle passing through all three vertices, and one inscribed circle (incircle) touching all three sides from inside. Regular polygons — squares, hexagons, octagons and so on — have both circles too. This calculator finds the circumradius R and inradius r, the areas and circumferences of both circles, and the area of the shape itself.
For a triangle, enter the three side lengths; the calculator uses Heron’s formula for the area, then R = abc ÷ 4K and r = K ÷ s. For a regular polygon, enter the number of sides and the side length; R and r (the apothem) follow from the central angle. It is useful for geometry homework and proofs, cutting the largest round table top from a triangular or hexagonal board (incircle), finding the smallest round stock a hexagonal or square bar can be turned from (circumcircle), and laying out polygons from a centre point.
For triangles it also reports the distance between the circumcentre and incentre from Euler’s formula d² = R(R − 2r), which shows Euler’s inequality R ≥ 2r — with equality only for an equilateral triangle.
With the default inputs, the circumradius (r) is 2.5. Change any value above to recalculate instantly.
How to use the circumcircle & incircle calculator
- 1Choose a triangle or a regular polygon.
- 2For a triangle, enter all three side lengths.
- 3For a regular polygon, enter the number of sides and the side length.
- 4Read the circumradius and inradius, plus both circle areas and circumferences.
Formula and method
For a triangle with sides a, b, c, the semi-perimeter is s = (a + b + c)/2 and Heron’s formula gives the area K. The circumradius follows from the extended law of sines (a ÷ sin A = 2R) combined with K = ½ab·sin C, which yields R = abc ÷ 4K. The incircle splits the triangle into three smaller triangles of height r, so K = r × s and r = K ÷ s. Euler’s theorem gives the distance between the two centres: d = √(R(R − 2r)).
A regular polygon with n sides of length s can be divided into n isosceles triangles meeting at the centre, each with apex angle 2π/n. Half of one of those triangles is a right triangle with opposite side s/2, giving R = (s/2) ÷ sin(π/n) and the apothem r = (s/2) ÷ tan(π/n). The polygon area is ½ × perimeter × apothem.
- a, b, c
- Triangle side lengths
- s
- Semi-perimeter (triangle) or side length (polygon)
- K
- Area
- R
- Circumradius
- r
- Inradius (apothem for regular polygons)
- n
- Number of sides
Worked examples
3-4-5 right triangle
Heron’s formula gives area 6 (semi-perimeter 6). R = 3 × 4 × 5 ÷ 24 = 2.5 — half the hypotenuse, as for every right triangle — and r = 6 ÷ 6 = 1. The centres are √(2.5 × 0.5) ≈ 1.118 apart.
Equilateral triangle with side 6
For an equilateral triangle R = side ÷ √3 ≈ 3.464 and r = side ÷ (2√3) ≈ 1.732, exactly half of R, and both circles share a centre.
Scalene triangle 7, 8, 9
With s = 12, the area is √(12 × 5 × 4 × 3) = √720 ≈ 26.83. Then R = 504 ÷ (4 × 26.83) ≈ 4.70 and r = 26.83 ÷ 12 = √5 ≈ 2.24.
Regular hexagon with 4 cm sides
A regular hexagon’s circumradius equals its side, 4 cm. The apothem is 4 ÷ (2 tan 30°) = 2√3 ≈ 3.46 cm, and the area is ½ × 24 × 3.46 ≈ 41.57 cm².
Frequently asked questions
How do you find the circumradius of a triangle?+
Use R = abc ÷ (4K), where K is the area (Heron’s formula from the three sides). For sides 5, 5, 6 the area is 12, so R = 150 ÷ 48 = 3.125.
How do you find the inradius of a triangle?+
Divide the area by the semi-perimeter: r = K ÷ s. For a 6-8-10 triangle, K = 24 and s = 12, so r = 2.
Where is the circumcentre of a triangle?+
Where the perpendicular bisectors of the sides meet. It lies inside an acute triangle, on the hypotenuse midpoint of a right triangle, and outside an obtuse triangle.
What is the circumradius of a regular hexagon?+
It equals the side length, because a regular hexagon splits into six equilateral triangles. A hexagon with 10 mm sides fits exactly in a 20 mm diameter circle.
What is the relationship between the circumradius and inradius?+
Euler’s inequality says R ≥ 2r for every triangle, with equality only when the triangle is equilateral. The distance between the two centres is √(R(R − 2r)).