Math Core

A chord is any segment that connects two points on a circle. Chords carry a lot of hidden structure: the center of the circle always sits on the perpendicular bisector of every chord, equal chords sit at equal distances from the center, and two crossing chords make angles you can find from arcs. These facts let you find a circle's radius from just a chord and a distance.

Chords and arcs

Every chord cuts off an arc (actually two, a minor arc and a major arc). In the same circle, the chord and its minor arc match up in size.

Congruent chords and arcs

In the same circle, or in congruent circles, two chords are congruent if and only if their minor arcs are congruent (which happens exactly when their central angles are congruent).

This works because the two radii and the chord form an isosceles triangle. If two chords are equal, the triangles are congruent by SSS, so the central angles are equal, and so are the arcs.

The perpendicular from the center

Draw a segment from the center straight down to a chord, meeting it at a right angle. Because the center is equally far from both endpoints of the chord, that segment lands exactly at the chord's midpoint.

Perpendicular bisector of a chord

If a diameter (or radius) is perpendicular to a chord, it bisects the chord and its arc.
The perpendicular bisector of any chord passes through the center of the circle.

The dashed segment OM is perpendicular to chord AB, so M is the midpoint of AB. Radius OB = 5, distance OM = 3, and half-chord MB = 4.

The figure shows the standard right triangle you get: the radius, the distance from the center to the chord, and half the chord. The radius is the hypotenuse:

$r^2 = d^2 + \left(\tfrac{c}{2}\right)^2,$

where $d$ is the distance from the center to the chord and $c$ is the chord length. The distance from a point to a chord is always measured along the perpendicular.

Worked example: Distance from the center

A circle has radius $13$ . A chord has length $24$ . How far is the chord from the center?

Half the chord is $12$ . In the right triangle, $13$ is the hypotenuse:

$d = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5.$

Worked example: Finding the radius

A chord of length $16$ is $6$ units from the center of a circle. Find the radius.

Half the chord is $8$ , so $r = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$ .

Common mistake

Use half the chord, not the whole chord. The perpendicular from the center splits the chord into two equal pieces, and only one of them is a leg of the right triangle.

Equal chords, equal distances

The right-triangle relationship shows that the chord length depends only on the distance from the center. That gives one more theorem.

Chords equidistant from the center

In the same circle, or in congruent circles, two chords are congruent if and only if they are the same distance from the center.

It also tells you that a chord closer to the center is longer. The longest chord of all, a diameter, has distance $0$ .

Angles formed by two chords

When two chords cross inside a circle, they form two pairs of vertical angles. Each angle "sees" an arc in front of it, and its vertical angle sees the arc on the opposite side.

Chords AC and BD cross at E. Angle AEB and its vertical angle CED intercept arcs AB and CD.

Angle formed by two chords

If two chords intersect inside a circle, each angle formed is half the sum of the arcs intercepted by the angle and its vertical angle:

$m\angle AEB = \dfrac{1}{2}\left(m\overset{\frown}{AB} + m\overset{\frown}{CD}\right).$

To see why, draw chord $\overline{AD}$ . Then $\angle AEB$ is an exterior angle of $\triangle AED$ , so it equals $m\angle EAD + m\angle EDA$ . Those are inscribed angles, equal to $\frac{1}{2}m\overset{\frown}{CD}$ and $\frac{1}{2}m\overset{\frown}{AB}$ .

Worked example: Crossing chords

Chords $\overline{AC}$ and $\overline{BD}$ intersect at $E$ . If $m\overset{\frown}{AB} = 70^\circ$ and $m\overset{\frown}{CD} = 110^\circ$ , find $m\angle AEB$ .

$m\angle AEB = \dfrac{1}{2}(70^\circ + 110^\circ) = \dfrac{1}{2}(180^\circ) = 90^\circ.$

The chords are perpendicular.

Tip

To find the center of a circle drawn on paper, draw two chords that aren't parallel and construct the perpendicular bisector of each. Both bisectors pass through the center, so they cross exactly there.