Circles: chords and angles

A cir­cle is the set of all points in a plane at the same dis­tance from a fixed point, its cen­tre. Because every radius is the same length, cir­cles are full of isosce­les and con­gru­ent tri­an­gles, and that is where all their prop­er­ties come from. This les­son proves the main facts about chords and the angles they make, end­ing with cyclic quadri­lat­er­als.

Words for parts of a cir­cle#

• A chord is a seg­ment join­ing two points on the cir­cle. The longest chords pass through the cen­tre; they are diam­e­ters, each twice the radius.
• An arc is a piece of the cir­cle between two points. A chord that is not a diam­e­ter cuts the cir­cle into a minor arc (the shorter) and a major arc (the longer).
• A seg­ment is the region between a chord and an arc; a sec­tor is the region between two radii and an arc.
• The angle a chord makes at the cen­tre is the angle between the radii to its ends.

Through­out, $O$ is the cen­tre of the cir­cle.

Equal chords#

The­o­rem. Equal chords of a cir­cle sub­tend equal angles at the cen­tre. Con­versely, chords that sub­tend equal angles at the cen­tre are equal.

Proof. Let $AB = CD$. In $\triangle AOB$ and $\triangle COD$: $OA = OC$ and $OB = OD$ (radii) and $AB = CD$. By SSS the tri­an­gles are con­gru­ent, so $\angle AOB = \angle COD$.

The per­pen­dic­u­lar from the cen­tre#

The­o­rem. The per­pen­dic­u­lar from the cen­tre to a chord bisects the chord. Con­versely, the line from the cen­tre to the mid­point of a chord is per­pen­dic­u­lar to it.

Proof. Let $OM \perp AB$. In right tri­an­gles $OMA$ and $OMB$: the hypotenuses $OA = OB$ (radii) and $OM$ is com­mon. By RHS they are con­gru­ent, so $AM = BM$.

This gives a right tri­an­gle with the radius as hypotenuse, half the chord as one leg and the dis­tance from the cen­tre as the other:

$$\displaystyle r^2 = \left(\tfrac{\text{chord}}{2}\right)^2 + d^2$$

A fur­ther result fol­lows: equal chords are equidis­tant from the cen­tre, and chords equidis­tant from the cen­tre are equal.

Exam­ple 1#

A chord of length $16$ cm is drawn in a cir­cle of radius $10$ cm. How far is it from the cen­tre?

Half the chord is $8$ cm, so $d = \sqrt{10^2 - 8^2} = \sqrt{36} = 6$ cm.

Exam­ple 2#

A chord is $5$ cm from the cen­tre of a cir­cle of radius $13$ cm. Find its length.

Half the chord is $\sqrt{13^2 - 5^2} = \sqrt{144} = 12$ cm, so the chord is $24$ cm.

Exam­ple 3#

$AB$ and $CD$ are par­al­lel chords of a cir­cle of radius $5$ cm, with $AB = 6$ cm and $CD = 8$ cm. Find the dis­tance between them.

The dis­tance of $AB$ from the cen­tre is $\sqrt{25 - 9} = 4$ cm, and of $CD$ is $\sqrt{25 - 16} = 3$ cm. Both per­pen­dic­u­lars lie on the same line through $O$ (per­pen­dic­u­lar to both chords).

If the chords are on oppo­site sides of the cen­tre, they are $4 + 3 = 7$ cm apart. If they are on the same side, they are $4 - 3 = 1$ cm apart.

Angles at the cen­tre and on the cir­cle#

The­o­rem. The angle sub­tended by an arc at the cen­tre is dou­ble the angle it sub­tends at any point on the remain­ing part of the cir­cle.

Proof (when $O$ lies inside the angle). Let arc $AB$ sub­tend $\angle AOB$ at the cen­tre and $\angle ACB$ at a point $C$ on the remain­ing part. Join $CO$ and extend it to a point $P$.

Tri­an­gle $OAC$ is isosce­les ($OA = OC$), so $\angle OAC = \angle OCA$. The exte­rior angle $\angle AOP$ equals the sum of these two, so $\angle AOP = 2\angle OCA$. In the same way $\angle BOP = 2\angle OCB$. Adding,

$$\angle AOB = \angle AOP + \angle BOP = 2(\angle OCA + \angle OCB) = 2\angle ACB$$

Three impor­tant con­se­quences:

1. Angle in a semi­cir­cle. If $AB$ is a diam­e­ter, $\angle AOB = 180°$, so any angle $\angle ACB$ with $C$ on the cir­cle is $90°$.
2. Angles in the same seg­ment are equal. All angles sub­tended by arc $AB$ at points of the remain­ing arc are half of the same $\angle AOB$, so they are equal.
3. Cyclic quadri­lat­er­als. A quadri­lat­eral with all four ver­tices on a cir­cle is cyclic. Its oppo­site angles add up to $180°$, because the two angles stand on the two arcs, whose cen­tral angles add to $360°$. Also, an exte­rior angle of a cyclic quadri­lat­eral equals the inte­rior oppo­site angle.

Exam­ple 4#

Arc $AB$ sub­tends $\angle AOB = 130°$ at the cen­tre. Find the angle it sub­tends at a point $C$ on the major arc, and at a point $D$ on the minor arc.

$\displaystyle \angle ACB = \frac{130°}{2} = 65°$. Then $ACBD$ is a cyclic quadri­lat­eral with $C$ and $D$ oppo­site, so $\angle ADB = 180° - 65° = 115°$.

Exam­ple 5#

$AB$ is a diam­e­ter of a cir­cle and $C$ is a point on the cir­cle with $\angle CAB = 35°$. Find $\angle ABC$.

$\angle ACB = 90°$ (angle in a semi­cir­cle), so $\angle ABC = 180° - 90° - 35° = 55°$.

Exam­ple 6#

$ABCD$ is a cyclic quadri­lat­eral with $\angle A = (2x + 4)°$ and $\angle C = (4x - 64)°$. Find both angles.

Oppo­site angles are sup­ple­men­tary: $6x - 60 = 180$, so $x = 40$. Then $\angle A = 84°$ and $\angle C = 96°$.

Prac­tice#

1. A chord of a cir­cle of radius $13$ cm is $24$ cm long. How far is it from the cen­tre?
2. A chord is $8$ cm from the cen­tre of a cir­cle of radius $17$ cm. Find its length.
3. A chord $10$ cm long is $12$ cm from the cen­tre. Find the radius.
4. Two par­al­lel chords of lengths $10$ cm and $24$ cm lie in a cir­cle of radius $13$ cm. Find the dis­tance between them if they are (a) on oppo­site sides of the cen­tre (b) on the same side.
5. Chords $AB$ and $CD$ of a cir­cle are equal, and $\angle AOB = 72°$. Find $\angle COD$.
6. Arc $AB$ sub­tends $100°$ at the cen­tre. Find the angle it sub­tends at a point on the major arc.
7. $C$ is a point on the major arc $AB$ and $\angle ACB = 40°$. Find $\angle AOB$ and $\angle OAB$.
8. $AB$ is a diam­e­ter and $C$ is on the cir­cle with $\angle ABC = 62°$. Find $\angle BAC$.
9. In cyclic quadri­lat­eral $ABCD$, $\angle A = 70°$ and $\angle B = 95°$. Find $\angle C$ and $\angle D$.
10. In cyclic quadri­lat­eral $ABCD$, side $BC$ is extended to $E$ and $\angle DCE = 105°$. Find $\angle A$.

Answers#