Circle and Related Theorems

Lessons

Senior 3 · Mathematics

Circle and Related Theorems

5 lessons

Lesson 2 of 5

Chords, and the angle facts every proof uses

25 min

Where is the middle of a round table?

A carpenter in Nyamirambo has made a round table top and must drill the hole for its single central leg. The centre is not marked, and measuring "by eye" leaves the table wobbling. With a ruler and a set square she can find the exact centre in two minutes, using a property of chords. This lesson proves that property, and first collects the angle facts you will quote as reasons in every circle proof of this unit.

The angle facts you already know

You met these facts in Senior 1. In circle geometry each line of working must name one of them, in brackets, as its reason.

Fact	Short reason to write
Angles on a straight line add up to $180^\circ$ .	(angles on a straight line)
Angles at a point add up to $360^\circ$ .	(angles at a point)
Vertically opposite angles are equal.	(vertically opposite angles)
The angles of a triangle add up to $180^\circ$ .	(angle sum of a triangle)
An exterior angle of a triangle equals the sum of the two interior opposite angles.	(exterior angle of a triangle)
The angles of a quadrilateral add up to $360^\circ$ .	(angle sum of a quadrilateral)
The base angles of an isosceles triangle are equal.	(base angles of an isosceles triangle)

Important

The fact that unlocks most circle problems: all radii of a circle are equal. So any triangle with two radii as sides, such as $\triangle OAB$ for a chord $A B$ , is isosceles, and its base angles are equal.

Worked example 1: an isosceles triangle from two radii

$A B$ is a chord of a circle with centre $O$ , and $\angle OAB = 34^\circ$ . Find $\angle AOB$ .

Solution

Statement	Reason
$O A = O B$	(radii)
$\angle OBA = \angle OAB = 34^\circ$	(base angles of an isosceles triangle)
$\angle AOB = 180^\circ - 34^\circ - 34^\circ = 112^\circ$	(angle sum of a triangle)

The perpendicular from the centre to a chord

Important

Chord theorem. The perpendicular from the centre of a circle to a chord bisects the chord: if $OM \perp AB$ , then $A M = M B$ .

Proof. In triangles $O M A$ and $O M B$ : $O A = O B$ (radii); $O M$ is common; $\angle OMA = \angle OMB = 90^\circ$ (given). So the triangles are congruent (RHS), and therefore $A M = M B$ .

Converse. The line from the centre to the midpoint of a chord is perpendicular to the chord. (Here the triangles are congruent by SSS, so the two angles at $M$ are equal; they are on a straight line, so each is $90^\circ$ .)

Consequences.

The perpendicular bisector of every chord passes through the centre.
In the right-angled triangle $O M A$ : $r^2 = d^2 + \left(\frac{1}{2}AB\right)^2$ , where $d = O M$ is the distance of the chord from the centre.
Equal chords are the same distance from the centre, and chords the same distance from the centre are equal. (The same $r$ and the same half-chord give the same $d$ .)

Worked example 2: finding the centre of the table

Mark any three points $P$ , $Q$ and $R$ on the edge of the table. Draw the chords $P Q$ and $Q R$ , and the perpendicular bisector of each.

The perpendicular bisector of $P Q$ passes through the centre (chord theorem). So does the perpendicular bisector of $Q R$ . Two different straight lines meet at only one point, so that point is the centre $O$ .

Worked example 3: a chord and its distance

(a) A chord is 5 cm from the centre of a circle of radius 13 cm. How long is it? (b) A chord 16 cm long lies in a circle of radius 10 cm. How far is it from the centre?

Solution (a) Half the chord: $\sqrt{13^2 - 5^2} = \sqrt{144} = 12$ cm (Pythagoras in $\triangle OMA$ ). The chord is $2 \times 12 = 24$ cm (the perpendicular from the centre bisects the chord).

(b) Half the chord is 8 cm (the perpendicular from the centre bisects the chord), so $d = \sqrt{10^2 - 8^2} = \sqrt{36} = 6$ cm (Pythagoras).

Worked example 4: two parallel chords

A circle has radius 5 cm. Two parallel chords are 6 cm and 8 cm long. Find the distance between them.

Solution 6 cm chord: $d = \sqrt{5^2 - 3^2} = 4$ cm. 8 cm chord: $d = \sqrt{5^2 - 4^2} = 3$ cm (chord theorem and Pythagoras each time).

The question does not say where the chords are, so there are two answers: 1 cm if they are on the same side of $O$ , and 7 cm if they are on opposite sides.

Worked example 5: water in a pipe

A horizontal pipe has an inner radius of 25 cm. The water surface inside it is 40 cm wide. How deep is the water, if it is less than half full?

Solution The water surface is a chord of 40 cm, so its half is 20 cm (the perpendicular from the centre bisects the chord). Its distance from the centre: $d = \sqrt{25^2 - 20^2} = \sqrt{225} = 15$ cm (Pythagoras). The depth is radius minus $d$ : $25 - 15 = 10$ cm. (If the pipe were more than half full, the depth would be $25 + 15 = 40$ cm.)

Warning

Use HALF the chord in Pythagoras. The right-angled triangle is $O M A$ , whose side is $\frac{1}{2}AB$ , not $A B$ .

"Perpendicular from the centre" is the condition. A perpendicular to a chord from any other point does not bisect it.

Tip

Every chord calculation is one right-angled triangle: radius (hypotenuse), half-chord and distance from the centre. Draw it, mark the right angle at the midpoint, and use Pythagoras.

Khan Academy: Perpendicular radius bisects chord. Watch the proof by congruent right-angled triangles (RHS; the video calls it RSH), the same proof as the chord theorem above.

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Practice

Give lengths to 3 significant figures where they are not exact. The last two are harder.

A chord 12 cm long lies in a circle of radius 10 cm. How far is it from the centre?
A chord is 5 cm from the centre of a circle of radius 7 cm. How long is it?
A chord 30 cm long is 8 cm from the centre. Find the radius.
$A B$ is a chord of a circle with centre $O$ , and $\angle OAB = 28^\circ$ . Find $\angle AOB$ , giving a reason for each step.
$A B$ is a chord with $\angle AOB = 100^\circ$ . Find $\angle OAB$ .
Two parallel chords, 10 cm and 24 cm long, lie in a circle of radius 13 cm. Find the two possible distances between them.
An arch is part of a circle. Its span (a chord) is 16 m, and its highest point is 4 m above the middle of the span. Find the radius of the circle.
Chords $A B$ and $C D$ of the same circle are equal. Prove that they are the same distance from the centre.

Show answersHide answers

$\sqrt{100 - 36} = 8$ cm.
$2\sqrt{49 - 25} = 2\sqrt{24} \approx 9.80$ cm.
$\sqrt{15^2 + 8^2} = \sqrt{289} = 17$ cm.
$\angle OBA = 28^\circ$ (base angles of an isosceles triangle, $O A = O B$ radii); $\angle AOB = 180^\circ - 56^\circ = 124^\circ$ (angle sum of a triangle).
$\angle OAB = \angle OBA$ (base angles, radii), so $\angle OAB = \frac{180^\circ - 100^\circ}{2} = 40^\circ$ .
Distances from the centre: $\sqrt{169 - 25} = 12$ cm and $\sqrt{169 - 144} = 5$ cm. Between the chords: $12 - 5 = 7$ cm (same side) or $12 + 5 = 17$ cm (opposite sides).
Let the radius be $r$ . The centre is $r - 4$ below the top, so $r^{2} = 8^{2} + (r - 4)^{2}$ , giving $0 = 64 - 8 r + 16$ and $r = 10$ m.
Let $M$ and $N$ be the feet of the perpendiculars from $O$ . Then $AM = \frac{1}{2}AB = \frac{1}{2}CD = CN$ (chord theorem), $O A = O C$ (radii), so $O M^{2} = O A^{2} - A M^{2} = O C^{2} - C N^{2} = O N^{2}$ (Pythagoras), and $O M = O N$ .

Sources and credits

Written by Ganzaa.
Video: Perpendicular radius bisects chord, Khan Academy, embedded from YouTube.

The diagrams, proofs, worked examples and practice items were written by Ganzaa; the table, pipe and arch measurements are illustrative.

1Where is the middle of a round table?

2The angle facts you already know

3The perpendicular from the centre to a chord

Where is the middle of a round table?

The angle facts you already know

The perpendicular from the centre to a chord