Circumscribing and inscribing circles

Hand a student a triangle and a compass and ask for "the circle that touches all three sides" and "the circle that passes through all three corners," and two very different points appear as the answer — even though both live inside or around the same triangle. One circle hugs the triangle from outside its edges, tangent to each side; the other stretches out to catch every vertex on its rim. Finding their centres by construction is not guesswork: both centres are loci, points defined by an equal-distance condition, and ICSE Class X asks you to build them with only a ruler and compass. This article extends the construction recipe with the numbers behind it. The circumcentre sits where the perpendicular bisectors of the sides meet, because that is the only point equidistant from all three vertices. The incentre sits where the angle bisectors meet, because that is the only point equidistant from all three sides. Along the way you get the standard formulas for the circumradius $R$ and the inradius $r$ , so you can check a construction numerically instead of trusting the ruler alone — and you get the neat shortcut for a regular hexagon, where the circumradius simply equals the side length.

The core insight

The circumcentre is the point equidistant from a triangle's three vertices (perpendicular bisectors of the sides), the incentre is the point equidistant from its three sides (bisectors of the angles), and both radii can be checked numerically from $R = \dfrac{abc}{4K}$ and $r = \dfrac{K}{s}$ , where $K$ is the triangle's area and $s$ its semi-perimeter.

The core idea

Circumscribing circle (circumcircle). A circle drawn around a triangle so that it passes through all three vertices is the circumcircle, and its centre is the circumcentre. Because the circumcentre is the same distance from every vertex, it must lie on the perpendicular bisector of every side — the perpendicular bisector of a segment is exactly the locus of points equidistant from the segment's two endpoints. Drawing the perpendicular bisectors of any two sides and marking where they cross gives the circumcentre directly; the third bisector automatically passes through the same point, which is a built-in accuracy check.

Inscribing circle (incircle). A circle drawn inside a triangle so that it just touches (is tangent to) all three sides is the incircle, and its centre is the incentre. Because the incentre is the same perpendicular distance from every side, it must lie on the bisector of every angle — the bisector of an angle is exactly the locus of points equidistant from the angle's two arms. Drawing the bisectors of any two angles and marking their intersection gives the incentre; the third angle's bisector will also pass through it.

Turning the picture into numbers. Once a triangle's three sides $a$ , $b$ , $c$ are known, its area $K$ can be found by Heron's formula, and that area unlocks both radii without any drawing at all:

K = \sqrt{s(s-a)(s-b)(s-c)}, \qquad s = \frac{a+b+c}{2}

where: $s$ is the semi-perimeter (half the perimeter) and $a, b, c$ are the three side lengths.

R = \frac{abc}{4K}, \qquad\qquad r = \frac{K}{s}

where: $R$ is the circumradius (centre-to-vertex distance of the circumcircle) and $r$ is the inradius (centre-to-side perpendicular distance of the incircle). These formulas let you predict the radius before you construct anything, then measure your drawn circle against the prediction.

Regular hexagon shortcut. A regular hexagon splits into six equilateral triangles meeting at its centre, so the distance from the centre to any vertex equals the side length itself: the circumradius of a regular hexagon equals its side, $R_{\text{hex}} = a$ . This is why a regular hexagon is constructed by stepping the compass around a circle at the same radius as the side — six such arcs close the hexagon exactly. Its incircle touches the midpoint of every side at the apothem distance, which is $r_{\text{hex}} = a\frac{\sqrt3}{2}$ , the height of one of those equilateral triangles.

Figure

A triangle with its circumcentre outside near one side and its incentre inside, showing perpendicular bisectors meeting at the circumcentre and angle bisectors meeting at the incentre

Derivation AWhy the incentre gives area equal to r times semi-perimeter

Show that if the incentre $I$ of a triangle with sides $a$ , $b$ , $c$ and area $K$ lies at perpendicular distance $r$ from every side, then $r$ must equal $K$ divided by the semi-perimeter $s = (a+b+c)/2$ , which is exactly the inradius formula used to predict a construction's radius before drawing it.

Step 1. Join the incentre $I$ to all three vertices $P$ , $Q$ , $R$ . This splits the original triangle $PQR$ into three smaller triangles, $\triangle PQI$ , $\triangle QRI$ , and $\triangle RPI$ , which together make up the whole triangle with no overlap.

Step 2. Each small triangle has one side of the original triangle as its base and $I$ as its opposite vertex. Because $I$ is the incentre, its perpendicular distance to every side is the same value $r$ — so $r$ is the height of all three small triangles, measured from $I$ down to whichever side is the base.

Step 3. The area of a triangle is half its base times its height, so the three small triangles have areas $\tfrac12 \cdot c \cdot r$ (base $PQ = c$ ), $\tfrac12 \cdot a \cdot r$ (base $QR = a$ ), and $\tfrac12 \cdot b \cdot r$ (base $RP = b$ ), using the standard convention that side $a$ is opposite vertex $P$ 's partner and so on.

Step 4. Adding the three areas gives the whole triangle's area: $K = \tfrac12 r(a+b+c)$ . Since the semi-perimeter is $s = (a+b+c)/2$ , the sum $a+b+c$ equals $2s$ , so $K = \tfrac12 r (2s) = rs$ . Dividing both sides by $s$ gives the inradius formula, $r = K/s$ — the same distance $r$ that the compass traces out for every side, confirmed algebraically rather than only by construction.

Worked example

Construct the circumcircle and estimate both special circles for $\triangle ABC$ with $AB = 6\ \text{cm}$ , $BC = 7\ \text{cm}$ , $CA = 5\ \text{cm}$ (the same triangle as the basic construction), and check the construction against the formulas.

Step 1 — Draw the triangle and construct the circumcentre. Draw $BC = 7\ \text{cm}$ , then locate $A$ with compass arcs of radius $6\ \text{cm}$ from $B$ and $5\ \text{cm}$ from $C$ . Construct the perpendicular bisectors of $AB$ and $BC$ ; they meet at the circumcentre $O$ . Measure $OA$ with the compass and draw the circumcircle — it should pass through $A$ , $B$ , and $C$ .

Step 2 — Compute the semi-perimeter and area. With $a = BC = 7$ , $b = CA = 5$ , $c = AB = 6$ , the semi-perimeter is $s = (7+5+6)/2 = 9\ \text{cm}$ . By Heron's formula, $K = \sqrt{9 \times 2 \times 4 \times 3} = \sqrt{216} = 6\sqrt6 \approx 14.7\ \text{cm}^2$ .

Step 3 — Predict the circumradius and compare. $R = \dfrac{abc}{4K} = \dfrac{7 \times 5 \times 6}{4 \times 14.7} = \dfrac{210}{58.8} \approx 3.57\ \text{cm}$ . Measure the constructed $OA$ with a ruler — it should read close to $3.6\ \text{cm}$ ; small differences of a millimetre or two are normal drawing error, not a sign the construction is wrong.

Step 4 — Predict the inradius for the same triangle. Using the derivation above, $r = K/s = 14.7/9 \approx 1.63\ \text{cm}$ . If you also constructed the incircle of this triangle (bisecting any two angles to find the incentre $I$ ), the perpendicular distance from $I$ to any side should measure close to $1.6\ \text{cm}$ . The circumradius is about $3.57\ \text{cm}$ and the inradius is about $1.63\ \text{cm}$ for this triangle — two very different numbers for two very different circles, exactly as the construction shows visually.

Competitive stretch

For a regular hexagon of side $5\ \text{cm}$ , the circumradius is simply $R_{\text{hex}} = 5\ \text{cm}$ (same as the side), while the inradius (apothem) is $r_{\text{hex}} = 5 \times \frac{\sqrt3}{2} \approx 4.33\ \text{cm}$ — notice how much closer the two hexagon radii are to each other than the triangle's $R \approx 3.57$ and $r \approx 1.63$ , because a hexagon is "rounder" than a scalene triangle.

Common traps

Confusing the circumcentre (perpendicular bisectors of sides) with the incentre (bisectors of angles) — swapping them gives a circle that neither passes through the vertices nor touches the sides.
Measuring the circumradius to the midpoint of a side instead of to a vertex; the circumcircle passes through vertices, so $R$ is always centre-to-vertex.
Measuring the inradius to a vertex instead of the perpendicular distance to a side; the incircle is tangent to the sides, so $r$ is always centre-to-side, measured perpendicularly.
Using $R = a$ (the hexagon shortcut) for a triangle, or using Heron's formula unnecessarily for a regular hexagon — each shortcut belongs to its own shape only.

Ask the tutor: stuck on whether a construction problem wants a circumcircle or an incircle, or want your Heron's-formula check walked through step by step? Ask the AI tutor.

Why does the circumcentre sometimes fall outside the triangle?

The circumcentre only has to be equidistant from the three vertices — nothing forces it to sit inside the triangle. For an acute triangle it falls inside, for a right triangle it falls exactly on the midpoint of the hypotenuse, and for an obtuse triangle it falls outside, on the far side of the longest side. The incentre, by contrast, is always inside the triangle, because it is defined by distances to the sides themselves.

Do I need Heron's formula every time I construct these circles?

No — for the actual ruler-and-compass construction you never calculate anything; you only draw bisectors and mark intersections. The formulas $R = abc/4K$ and $r = K/s$ are a separate check you can use afterward, by measuring the drawn radius and comparing it with the predicted number, or in problems that give you the sides and ask for the radius directly without drawing.

Why is the hexagon's circumradius equal to its side length?

A regular hexagon can be cut into six identical equilateral triangles meeting at the centre, because each central angle is $360^\circ/6 = 60^\circ$ and the two sides from the centre to adjacent vertices are equal, forcing an equilateral triangle. In an equilateral triangle every side is equal, so the centre-to-vertex distance (which is the hexagon's circumradius) equals the shared side length.

Circumscribing and inscribing circles

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads