Arcs, chords and central angles

Geometry, grade 10 (5 units) Lesson 19 of 26Learning

Arcs, chords and central angles

The parts of a circle

A circle is the locus of the points at a given distance from a given point: the point is the centre, the distance the radius.

Radius – a segment from the centre to a point on the circle, like $\displaystyle OA$ .
Chord – a segment joining two points on the circle, like $\displaystyle BC$ .
Diameter – a chord through the centre, like $\displaystyle DE$ ; it is two radii long.
Arc – a part of the circle between two of its points; two points cut the circle into two arcs.
Central angle – an angle whose vertex is the centre and whose arms are radii, like $\displaystyle \angle BOC$ . It stands on the arc inside it, and the arc is measured by it: an arc of $\displaystyle 80^{\circ}$ is the arc of a central angle of $\displaystyle 80^{\circ}$ .

Equal central angles and equal chords

Theorem: In a circle, equal central angles stand on equal arcs and on equal chords.

Given: $\displaystyle O$ is the centre of the circle; $\displaystyle {\angle AOB}={\angle COD}$ .
To prove: $\displaystyle {AB}={CD}$ .

The proof below shows that the triangles $\displaystyle AOB$ and $\displaystyle COD$ are congruent (SAS) – their sides from $\displaystyle O$ are radii:

	Claim	Reason
1	$\displaystyle {OA}={OC}$	Radii of the circle
2	$\displaystyle {OB}={OD}$	Radii of the circle
3	$\displaystyle {\angle AOB}={\angle COD}$	Given
4	$\displaystyle \triangle AOB\cong\triangle COD$	(1), (2), (3); the SAS congruence theorem
5	$\displaystyle {AB}={CD}$	(4); corresponding sides of congruent triangles

The converse holds too: in a circle, equal chords stand on equal central angles and on equal arcs (Exercise 1). So in one circle – or in two circles with equal radii – equal central angles, equal arcs and equal chords go together.

Exercise 1

$\displaystyle O$ is the centre of the circle, and $\displaystyle {AB}={CD}$ .
Prove: $\displaystyle {\angle AOB}={\angle COD}$ .

The perpendicular from the centre to a chord

Theorem: The perpendicular from the centre of a circle to a chord bisects the chord, the central angle on the chord, and the arc on the chord.

Given: $\displaystyle AB$ is a chord of the circle with centre $\displaystyle O$ ; $\displaystyle OM\perp AB$ , and $\displaystyle M$ is on $\displaystyle AB$ .
To prove: $\displaystyle {AM}={MB}$ and $\displaystyle {\angle AOM}={\angle BOM}$ .

The proof below shows that the right-angled triangles $\displaystyle OMA$ and $\displaystyle OMB$ are congruent, by the fourth congruence theorem – two sides and the angle opposite the longer of them, here the right angle:

	Claim	Reason
1	$\displaystyle {OA}={OB}$	Radii of the circle
2	$\displaystyle {\angle OMA}={\angle OMB}$	$\displaystyle OM\perp AB$ : both angles are right
3	$\displaystyle {OM}={OM}$	A common side
4	$\displaystyle \triangle OMA\cong\triangle OMB$	(1), (2), (3); two sides and the angle opposite the longer of them
5	$\displaystyle {AM}={MB}$	(4); corresponding sides of congruent triangles
6	$\displaystyle {\angle AOM}={\angle BOM}$	(4); corresponding angles of congruent triangles

The central angles $\displaystyle AOM$ and $\displaystyle BOM$ are equal, so they stand on equal arcs: $\displaystyle OM$ , extended to the circle, bisects the arc $\displaystyle AB$ too.

Exercise 2

The converse theorem: the segment from the centre to the midpoint of a chord is perpendicular to the chord.
$\displaystyle M$ is the midpoint of the chord $\displaystyle AB$ of the circle with centre $\displaystyle O$ .
Prove: $\displaystyle OM\perp AB$ .

The distance of a chord from the centre

The distance of a chord from the centre is the perpendicular $\displaystyle d$ from the centre to it. The perpendicular bisects the chord, so half the chord, $\displaystyle d$ and the radius make a right-angled triangle:

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

\displaystyle =

\displaystyle R^{2}

Equal chords are at equal distances from the centre, and chords at equal distances are equal (Exercise 5).
The longer the chord, the nearer it is to the centre: $\displaystyle {d^{2}}={R^{2}-\left(\frac{AB}{2}\right)^{2}}$ gets smaller as $\displaystyle AB$ grows. The longest chord – the diameter – passes through the centre, $\displaystyle {d}={0}$ .

Example: in a circle of radius $\displaystyle 13$ , a chord is $\displaystyle 24$ long. Half of it is $\displaystyle 12$ , so $\displaystyle {d}={\sqrt{13^{2}-12^{2}}}={\sqrt{25}}={5}$ .

Exercise 3

Find.

The radius of a circle is

\displaystyle 10

, and a chord is

\displaystyle 6

from the centre. How long is the chord?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

A chord

\displaystyle 16

long is

\displaystyle 15

from the centre. What is the radius?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 4

In a circle of radius $\displaystyle 5$ , two parallel chords are $\displaystyle 6$ and $\displaystyle 8$ long, on the two sides of the centre.
Find the distance between the chords.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 5

$\displaystyle AB$ and $\displaystyle CD$ are equal chords of the circle with centre $\displaystyle O$ .
$\displaystyle OM\perp AB$ and $\displaystyle ON\perp CD$ .
Prove: $\displaystyle {OM}={ON}$ .

Exercise 6

A circle is drawn, but its centre is not marked.
Describe how to find the centre with a ruler and compasses, and explain why it works.
(A hint: draw two chords that are not parallel.)

Formulas: geometry, grade 10 (5 units)

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Introduction to analytic geometry

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{(x_1-x_2)^{2}+(y_1-y_2)^{2}}$
The midpoint of the segment
$\displaystyle M\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$
The slope of a line through two points
$\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ $\displaystyle x_1\neq x_2$
The line of slope $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$
$\displaystyle y-y_1$ $\displaystyle =$ $\displaystyle m(x-x_1)$
The explicit equation of a line – $\displaystyle m$ the slope, $\displaystyle b$ where it crosses the $\displaystyle y$ -axis
$\displaystyle y=mx+b$
The general equation of a line
$\displaystyle Ax+By+C=0$
Parallel lines
$\displaystyle m_1$ $\displaystyle =$ $\displaystyle m_2$
Perpendicular lines
$\displaystyle m_1\cdot m_2$ $\displaystyle =$ $\displaystyle -1$
A canonical circle – its centre at the origin, $\displaystyle R$ its radius
$\displaystyle x^{2}+y^{2}=R^{2}$
The circle with centre $\displaystyle (a,b)$ and radius $\displaystyle R$
$\displaystyle (x-a)^{2}+(y-b)^{2}=R^{2}$

Midsegments and the meeting point of the medians

A triangle’s midsegment – $\displaystyle DE$ is parallel to $\displaystyle BC$
$\displaystyle DE$ $\displaystyle =$ $\displaystyle \frac{1}{2}BC$
A trapezoid’s midsegment – $\displaystyle a$ , $\displaystyle b$ the bases
$\displaystyle EF$ $\displaystyle =$ $\displaystyle \frac{a+b}{2}$
The meeting point of the medians $\displaystyle G$ divides the median $\displaystyle AD$ in the ratio $\displaystyle 2:1$
$\displaystyle AG$ $\displaystyle =$ $\displaystyle 2GD$
Pythagoras’ theorem and its converse: the triangle is right-angled, the right angle opposite $\displaystyle c$
$\displaystyle a^{2}+b^{2}=c^{2}$

Proportion and similarity

A property of a proportion
$\displaystyle \frac{a}{b}=\frac{c}{d}\ \Rightarrow\ \frac{a+b}{b}=\frac{c+d}{d}$
Thales’ theorem – $\displaystyle DE\parallel BC$
$\displaystyle \frac{AD}{DB}$ $\displaystyle =$ $\displaystyle \frac{AE}{EC}$
The extended Thales theorem – $\displaystyle DE\parallel BC$
$\displaystyle \frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$
The angle bisector theorem – $\displaystyle AD$ bisects the angle $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$
The similarity ratio – corresponding sides
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$
Heights, medians, perimeters, the radii of the circumscribed and of the inscribed circles
$\displaystyle \frac{h_1}{h_2}=\frac{m_1}{m_2}=\frac{P_1}{P_2}=\frac{R_1}{R_2}=\frac{r_1}{r_2}=k$
The ratio of the areas
$\displaystyle \frac{S_1}{S_2}$ $\displaystyle =$ $\displaystyle k^{2}$

The circle

An inscribed angle is half the central angle on the same arc
$\displaystyle \angle ACB$ $\displaystyle =$ $\displaystyle \frac{1}{2}\angle AOB$
An inscribed angle on the diameter $\displaystyle AB$
$\displaystyle \angle ACB=90^{\circ}$
A quadrilateral $\displaystyle ABCD$ inscribed in a circle – opposite angles
$\displaystyle \angle A+\angle C$ $\displaystyle =$ $\displaystyle 180^{\circ}$
The angle between the tangent $\displaystyle AT$ and the chord $\displaystyle AB$ equals an inscribed angle on $\displaystyle AB$ from its other side
$\displaystyle \angle BAT$ $\displaystyle =$ $\displaystyle \angle ACB$
Two tangents from the point $\displaystyle P$
$\displaystyle PA=PB$
A quadrilateral $\displaystyle ABCD$ circumscribing a circle
$\displaystyle AB+CD$ $\displaystyle =$ $\displaystyle BC+AD$
The chords $\displaystyle AB$ and $\displaystyle CD$ meet at the point $\displaystyle E$
$\displaystyle AE\cdot EB$ $\displaystyle =$ $\displaystyle CE\cdot ED$
Two secants from the point $\displaystyle P$
$\displaystyle PA\cdot PB$ $\displaystyle =$ $\displaystyle PC\cdot PD$
A tangent $\displaystyle PT$ and a secant $\displaystyle PAB$ from the point $\displaystyle P$
$\displaystyle PT^{2}$ $\displaystyle =$ $\displaystyle PA\cdot PB$