Chords in circles
Master chords in circles for Year 11 Specialist Mathematics in Queensland (QCAA). A chord joins two points on a circle, and a few exact rules — the perpendicular from the centre bisects a chord, equal chords are equidistant from the centre, and the intersecting-chords theorem — let you find unknown lengths.
You will learn to drop the perpendicular from the centre, build the right triangle of radius, distance and half-chord, and apply Pythagoras and the chord and secant products — the circle-geometry groundwork for the angle theorems and formal proofs later in the topic.
Theory
Chords in circles obey a small set of exact rules used throughout Year 11 Specialist Mathematics (QCAA, Queensland). The perpendicular from the centre bisects a chord, equal chords are equidistant from the centre, and where two chords cross the products of their segments are equal. This page shows how each rule gives a right triangle or an equation you solve for an unknown length.
A chord is a line segment joining two points on a circle. The longest chord passes through the centre and is a diameter. A radius joins the centre \(O\) to a point on the circle.
Perpendicular from the centre. The perpendicular drawn from the centre of a circle to a chord bisects the chord: it meets the chord at its midpoint \(M\), so \(AM=MB\). The converse also holds — the line from the centre to the midpoint of a chord is perpendicular to it.
Equal chords. Chords of equal length are the same perpendicular distance from the centre, and conversely chords equidistant from the centre are equal in length.
Intersecting chords. When two chords meet at a point \(P\) inside a circle, the products of the two segments of each chord are equal: \(AP\times PB = CP\times PD\). For a tangent and a secant from an external point \(P\), the tangent squared equals that same product: \(PT^2=PA\times PB\).
Dropping the perpendicular from the centre \(O\) to a chord makes a right triangle whose hypotenuse is the radius \(r\), one leg is the distance \(d\) from the centre, and the other leg is half the chord:
Rearranged, the full chord length is
Where two chords cross at an interior point \(P\), and for a tangent \(PT\) with a secant \(PAB\) from an external point:
Finding an unknown length
- Draw and drop the perpendicular: mark the centre \(O\), the chord, and the perpendicular \(OM\) from the centre to the chord's midpoint \(M\).
- Form the right triangle: the radius \(r\) is the hypotenuse, the distance \(d\) from the centre and the half-chord \(AM\) are the legs.
- Apply the rule: use Pythagoras \(r^{2}=d^{2}+AM^{2}\) for a chord, radius or distance; use \(AP\times PB=CP\times PD\) (or \(PT^{2}=PA\times PB\)) for intersecting chords or a tangent and secant.
- Finish: double the half-chord for the whole chord, reject any negative root, and leave surds exact.
The distance, the half-chord and the radius form a right triangle; find the half-chord by Pythagoras, then double it.
| \(AM^2\) | \(=\) | \(OA^2 - OM^2\) |
| \(AM^2\) | \(=\) | \(17^2 - 8^2\) |
| \(AM^2\) | \(=\) | \(289 - 64\) |
| \(AM^2\) | \(=\) | \(225\) |
| \(AM\) | \(=\) | \(15\) |
| \(AB\) | \(=\) | \(2\times 15\) |
| \(=\) | \(30\) |
The chord is \(30\) cm long.
The perpendicular from the centre bisects the chord, so the half-chord is \(20\); use Pythagoras for the distance.
| \(AM\) | \(=\) | \(\dfrac{40}{2}=20\) |
| \(OM^2\) | \(=\) | \(OA^2 - AM^2\) |
| \(OM^2\) | \(=\) | \(25^2 - 20^2\) |
| \(OM^2\) | \(=\) | \(625 - 400\) |
| \(OM^2\) | \(=\) | \(225\) |
| \(OM\) | \(=\) | \(15\) |
The distance is \(15\) cm.
By the intersecting-chords theorem the products of the two segments of each chord are equal.
| \(AP\times PB\) | \(=\) | \(CP\times PD\) |
| \(6\times 4\) | \(=\) | \(8\times PD\) |
| \(24\) | \(=\) | \(8\,PD\) |
| \(PD\) | \(=\) | \(\dfrac{24}{8}\) |
| \(=\) | \(3\) |
\(PD=3\) cm.
Find the half-chord by Pythagoras, simplify the surd, then double it.
| \(AM^2\) | \(=\) | \(9^2 - 5^2\) |
| \(AM^2\) | \(=\) | \(81 - 25\) |
| \(AM^2\) | \(=\) | \(56\) |
| \(AM\) | \(=\) | \(\sqrt{56}\) |
| \(=\) | \(2\sqrt{14}\) | |
| \(AB\) | \(=\) | \(2\times 2\sqrt{14}\) |
| \(=\) | \(4\sqrt{14}\) |
The chord is \(4\sqrt{14}\) cm long.
Common pitfalls
Frequently asked questions
What is a chord of a circle?
A chord is a straight line segment joining two points on the circle. The longest chord passes through the centre and is called a diameter.
Does the perpendicular from the centre bisect a chord?
Yes. The perpendicular drawn from the centre of a circle to a chord always meets the chord at its midpoint, so the two halves \(AM\) and \(MB\) are equal.
How do you find the length of a chord from the radius and its distance from the centre?
Use the right triangle formed by the radius, the distance \(d\) and half the chord: \(AM=\sqrt{r^{2}-d^{2}}\), then double it, so the chord \(=2\sqrt{r^{2}-d^{2}}\).
What is the intersecting chords theorem?
When two chords cross at a point \(P\) inside a circle, the products of their segments are equal: \(AP\times PB=CP\times PD\).
How do you find the distance from the centre to a chord?
Halve the chord, then use Pythagoras with the radius: \(d=\sqrt{r^{2}-AM^{2}}\), where \(AM\) is the half-chord.
Are equal chords the same distance from the centre?
Yes. Equal chords are equidistant from the centre, and conversely chords the same distance from the centre are equal in length.