Chord properties

Draw any straight line across a circle and, where it crosses the circle at two points, you get a chord. Circles enforce a strict discipline on their chords: draw the perpendicular from the centre to a chord and it always lands exactly on the chord's midpoint, whichever chord you pick. That single symmetry unlocks a small, powerful toolkit — a way to find how far a chord sits from the centre, a way to know two chords are equal just by comparing their distances from the centre, and a way to pin down the one circle passing through three scattered points. This section makes that toolkit precise: the perpendicular-bisector link between a chord and the centre, the equal-chords-equal-distance rule, and the three-non-collinear-points-one-circle result, all built from nothing more than congruent triangles and Pythagoras theorem.

The core insight

A radius drawn perpendicular to a chord always bisects it, so the distance $d$ from the centre, the half-chord $\tfrac12 AB$ , and the radius $r$ form a right triangle with $r^2 = d^2 + \left(\tfrac12 AB\right)^2$ , and the same idea shows equal chords are equidistant from the centre and that three non-collinear points fix exactly one circle.

The core idea

Let $O$ be the centre of a circle of radius $r$ , and let $AB$ be a chord that is not a diameter.

1. Bisector from the centre means perpendicular, and back again. If $M$ is the midpoint of $AB$ , then $OM \perp AB$ : triangles $OAM$ and $OBM$ share $OM$ , have equal radii $OA = OB$ , and equal halves $AM = MB$ , so they are congruent by SSS and the two angles at $M$ (which are supplementary) must each be $90^\circ$ . The converse also holds and is given without proof in the syllabus: the perpendicular dropped from the centre onto a chord always bisects that chord, $OM \perp AB \implies AM = MB$ . These two directions of the same fact are the single most-used tool in this topic.

2. Equal chords sit at equal distances, and back again. If two chords of the same circle are equal in length, their perpendicular distances from the centre are equal; and conversely, chords that lie at equal distances from the centre are equal in length (also given without proof). Both directions follow from the same right triangle described below, since a chord's length and its distance from the centre are locked together once the radius is fixed.

3. Three non-collinear points fix one circle. Given three points not all on one straight line, exactly one circle passes through all three. Join the points to form a triangle, draw the perpendicular bisector of any two sides, and their intersection is the unique centre — because every point on a perpendicular bisector of a segment is equidistant from that segment's two ends, so the intersection point is equidistant from all three original points and can serve as a centre. The computational engine behind all of this is a single right triangle. Drop the perpendicular $OM$ from the centre to chord $AB$ , meeting it at midpoint $M$ . Since $OM \perp AB$ , triangle $OMA$ has a right angle at $M$ , so by Pythagoras theorem:

r^2 = d^2 + \left(\frac{AB}{2}\right)^2

where: $r = OA$ is the radius, $d = OM$ is the perpendicular distance of the chord from the centre, and $AB$ is the chord's full length (so $\tfrac12 AB = MA$ is the half-chord). Given any two of $r$ , $d$ , $AB$ , this one equation solves for the third.

Figure

A circle with centre O and chord AB whose midpoint M is joined to O by a perpendicular, forming a right triangle with legs d and half of AB and hypotenuse r

Derivation AChord-distance-radius relation from Pythagoras theorem

The goal is to derive the relation $r^2 = d^2 + \left(\tfrac12 AB\right)^2$ that links a chord's length, its perpendicular distance from the centre, and the circle's radius, starting only from the perpendicular-bisects-chord property and Pythagoras theorem, so that any one of the three quantities can be found once the other two are known.

Step 1. Draw the chord $AB$ in a circle of centre $O$ and radius $r$ , and drop the perpendicular $OM$ from $O$ onto $AB$ , meeting it at $M$ . By the perpendicular-from-centre-bisects-chord property, $M$ is the midpoint of $AB$ , so $MA = MB = \tfrac12 AB$ .

Step 2. Because $OM \perp AB$ , the angle at $M$ in triangle $OMA$ is $90^\circ$ , so $OMA$ is a right triangle with legs $OM = d$ and $MA = \tfrac12 AB$ , and hypotenuse $OA = r$ (a radius).

Step 3. Apply Pythagoras theorem to triangle $OMA$ : the square on the hypotenuse equals the sum of the squares on the two legs, so $OA^2 = OM^2 + MA^2$ , which is exactly $r^2 = d^2 + \left(\tfrac12 AB\right)^2$ .

Step 4. Read the relation both ways. Given $r$ and $d$ , the chord length is $AB = 2\sqrt{r^2 - d^2}$ . Given $r$ and $AB$ , the distance is $d = \sqrt{r^2 - \left(\tfrac12 AB\right)^2}$ . Because $r$ is fixed for a given circle, a larger chord forces a smaller $d$ and vice versa — chords closer to the centre are longer, and the longest possible chord, the diameter, has $d = 0$ .

Worked example

A chord of length $16\text{ cm}$ is drawn in a circle of radius $10\text{ cm}$ . Find the distance of the chord from the centre.

Step 1. Let $M$ be the midpoint of chord $AB$ , so $MA = \tfrac12 \times 16 = 8\text{ cm}$ , using the perpendicular-bisects-chord property. Let $d = OM$ be the unknown distance and $OA = r = 10\text{ cm}$ .

Step 2. Apply $r^2 = d^2 + \left(\tfrac12 AB\right)^2$ to the right triangle $OMA$ : $10^2 = d^2 + 8^2$ , so $100 = d^2 + 64$ , giving $d^2 = 36$ and $d = 6\text{ cm}$ .

The chord is 6 cm from the centre.

Competitive stretch

In the same circle, two chords $PQ$ and $RS$ are each found to lie $5\text{ cm}$ from the centre $O$ , and $PQ = 24\text{ cm}$ . Find $RS$ and check what radius this implies.

Step 1. Since $PQ$ and $RS$ are equidistant from the centre ( $5\text{ cm}$ each), the "chords equidistant from the centre are equal" rule gives $RS = PQ = 24\text{ cm}$ directly, with no calculation needed.

Step 2. To check the radius, apply $r^2 = d^2 + \left(\tfrac12 PQ\right)^2$ with $d = 5\text{ cm}$ and $\tfrac12 PQ = 12\text{ cm}$ : $r^2 = 5^2 + 12^2 = 25 + 144 = 169$ , so $r = 13\text{ cm}$ . This is a different circle from the first part (radius $13\text{ cm}$ , not $10\text{ cm}$ ), and $RS = 24\text{ cm}$ .

Common traps

Applying the bisecting property to a diameter and expecting a special right angle — every line through the centre passes through a diameter, so the perpendicular-bisector content is only meaningful for genuine, non-diameter chords.
Using the full chord length instead of the half-chord in $r^2 = d^2 + \left(\tfrac12 AB\right)^2$ — always halve $AB$ before applying Pythagoras theorem.
Assuming equal chords must pass through the centre — equal chords are only equidistant from the centre; a chord through the centre itself is specifically a diameter.
Forgetting "not collinear" in the three-points rule — three points on a single straight line never determine a circle, since their perpendicular bisectors run parallel and never meet.

Ask the tutor: Want to work through more chord-length, distance, or radius problems using $r^2 = d^2 + \left(\tfrac12 AB\right)^2$ , or see the three-points-determine-a-circle construction step by step? Ask the AI tutor.

Why is the perpendicular-bisects-chord property given without proof but the other direction is proved?

Both directions are true, but only one needs the SSS congruence argument to be built from scratch in this syllabus: starting from " $M$ is the midpoint" and proving $OM \perp AB$ uses congruent triangles $OAM$ and $OBM$ cleanly. The converse — starting from " $OM \perp AB$ " and proving $M$ is the midpoint — needs a slightly different congruence setup that the ICSE Class 9 syllabus chooses to state as a fact rather than derive. You are expected to use both directions freely in problems even though only one is proved.

What is the actual difference between a chord and a diameter?

A chord is any line segment joining two points on a circle, full stop — a diameter is simply the special case of a chord that happens to pass through the centre. Every diameter is a chord, but not every chord is a diameter. The distinction matters here because the perpendicular-bisector property, while technically true for a diameter too, is only interesting for a general chord: a diameter always passes through the centre, so its "distance from the centre" is trivially zero, and there is nothing new to prove.

Why do we need three points to fix a circle instead of two?

Two points alone are not enough because infinitely many circles can pass through any two given points — their centres can lie anywhere along the perpendicular bisector of the segment joining them, since every point on that bisector is equidistant from the two given points. A third point, not on the same line as the first two, forces a second perpendicular bisector; the two perpendicular bisectors meet at exactly one point, which is the only centre equidistant from all three points, and hence the only circle through all three.

Chord properties

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads