## CIRCLE

**DEFINITION AND PROPERTIES OF A CIRCLE**

A circle can be defined in two ways.

**A circle**: Is a closed path curve all points of which are equal-distance from a fixed point called centre OR

– Is a locus at a point which moves in a plane so that it is always of constant distance from a fixed point known as a centre.

O – Is called the centre of the circle.

OP – Is the line segment is the radius.

AB – Is the line segment of diameters of the circle.

O – (Centre) – Is a fixed point of circle.

OP (Radius) – Is the constant distance from the centre to any point on a circumstance of a circle.

AB (Diameter) – Is a line segment which passes through the centre of a circle.

A circumference – Is a length of a locus which moves around the centre.

Diameter =2 x Radius

D = 2r

Hence the diameter of a circle is equal to two times radius.

RS (Secant) – Is a line segment whose points are on the circle.

BOC – Is called a central angle

(PTQ) – A Segment – Is the part of a circular region included within the chord and its arc.

(COB) Sector – Is the part of a circular region bounded by two radii and an arc.

**CENTRAL ANGLE**

Consider a circle of radius r, length of arc l, subtending a central angle.

– The length of the circumference C of the circle is C = 2πr. This means that the length of the arc intercept by a central angle 360º is 2πr.

– The length of an arc is proportional to the measure of the central angle of the central angle. Thus if the central angle is 1

**Example 1.**

An arc subtends an angle of 20^{0 }at the centre of a circle of radius 25m.

Find the length of the arc.

Length of an arc is given by

= Length of an arc = 8.72m

An arc of length 5cm subtends 50º at the centre of a circle. What is the radius of the circle.

**Data **

θ=50º

Length = 5 cm

Radius = required

Length of an arc is equal to

=5.73cm

**Example 3**

A circular running track has radius 50m. A sprinter runs 100m along the track.

Through what angle has she turned?

Radius = 50m

Length of an arc = 100m

θ= Required

Length at an arc is equal to

The central angle is 114 .650

** Questions:**

1. An arc subs tends 25 at the centre of a circle of radius 40m. What is the length of the arc?

**Data**

Length of an arc = required

Radius = 40m

θ=- 25

Length at an arc is equal to

The length at an arc = 17.44m

2. An arc of length 17cm forms a circle of radius 40cm what angle does the arc subtend?

**Data :-**

Length of an arc = 17cm

Radius = 40cm

θ= Required

Length of arc is equal to

The arc subtends 24.330

3. An arc of length 16m subtends 400 at the centre of the circle. What is the radius at the circle?

Data

Length an arc = 16m

Radius = Required

θ= 40^{0}

Length of arc is equal to

The radius = 22.85m

**RADIAN MEASURE**

Angles can also be used to measure the amount of turning. Turns of a minute hand of a clock and a wheel can be measured in both angles and radians. Example a minute hand of clock turns through an angle of 90º 0f ½ π radians between noon and 12:15pm as shown in the figure.

– The tip of the hand has covered a distance of ½ π radians.

From noon to 12:45 pm, the hrs turned through an angle of 270ºor 3/2πr radians. The angle 270º is reflex angle.

– One complete turn at hand clock represents an angle of 360º or 2π radians.

– Measures of angles more than 360^{0} or 2 π radius can be obtained if hand of a clock measures more than one complete turn. Example from noon to 1:15 pm, the hand has turned through 1 ¼ turns. Now one turn is 360^{0 }or 20 radians.

– 5/4 turns or 3600 x 5/4 0r 2π x 5/4 radians which reduce to 450º 0r 5/2π radians.

– There fore from noon to 1:15 pm the hand turns through 450º or 5/2π radians.

**Questions **

1. Give the size in degree at an angle through which a minute hand of a clock has returned between noon and the following times.

(a) 12:40

Solution:

1min = 6

40min =?

X= 40min x 6

X = 240

(b) 3:00

** Solution**

1hour = 360

3 hour =?

X = 3 hour x 360^{0}

X = 1080^{0}

(c) 9:00

** Solution**

1hour = 360^{0}

9 hour =? X

X = 9 x 360^{0}

X = 3240^{0}

2. Give the size in radians at angles through which the minute hand of a clock has turned between noon and the following times.

(a) 12:20pm

** Solution**

1min = 6^{0}

20 min =? X

X= 20min x 6

X = 120^{0}

πrad = 180º

? = 120º

x = π rad x 120

X = 2/3 π rad

(b) 2:15

** Solution**

1hour = 60 min

2 hour =? X

x =60 x 15min

X =120

120 x 6º=720º

= 720^{0 }+90^{º
} = 810º

(c) 24:00 noon

**Solution**

1 hour = 360^{0}

24hrs = ?

X = 360^{0} x 20

= 8640^{0}

πrad = 180^{0}

?x = 8640^{0}

x = 8640^{0} x πrad

X = 48πrad

**RADIAN MEASURE**

– The relation between the arc length l, the central angle θ, and the radius r, can be used to compare the measurement of an angle in radius with the measurements in degree.

Circumference of the circle for the given radian C = 2πr. Circumference sector

But C = length at an arc

In abbreviation is written as:-

S = is radian measured of an angle

θ= is the degree measure

θ= 17 .19^{0}

The angle in degree is 17. 19^{0}

** Class Activity:-**

1.Find the degree of each of the following :-

(i) 3/2π

**Solution:-**

= 90^{0 } x 3

= 3/2π = 270^{0}

(ii)¾π

**Solution**

¾ π = 135^{0}

**Example 1:**

Find in radian as multiple of π for each of the following degrees.

(a) 315^{0 } (b) 240^{0}

Solution:

(b) 240^{0}

Solution:-

2. Change the following radians into degree

(a) 0.3 (b) 5

** solution
**

**Class activity.**

1. Find the degree of each of the following

^{i) }3/2π

Solution.

ii 3/4 π

Solution

iii) 2 π

Solution

2.Find the radians multiple of the π following

(i)80°

**Solution**

ii) 215º

**Solution **

iii) 60^{0}

**Solution**

3. (i) Change the radians into the degree of 0.3

∴The angle the degree = 54^{0}

(ii) 5

**Solution**

The angle degree = 900^{0}

**ANGLES IN CYCLIC QUADRILATERAL**

These are four angles whose vertices are lying on the circumference of a circle.

The angles p, q, r, and t are called cyclic angles in a quadrilateral ABCD, q and t, p and r are opposite angles.

1. **THEOREM:**

The opposite angles of a cyclic quadrilateral are supplementary (add up to 180^{0}).

Given; A quadrilateral SPQR inscribed in a circle centered at 0

Required to prove: x+ y = 180^{0}

Constuction; join OR and OP

Proof: in the above figure

a = Zx (angle s on a circle) PQR)

b = Zy (angle on arc PSR)

a+b = Zx + 2y but a+ b = 360º

360^{0} = 2 (x+y ) divide by 2 both sides

x+y = 180^{0}

∴x+ y = 180º hence proved.

**Example: **

Find the size of each lettered angle.

The opposite angles A and C, B and D

:. x +83^{0} = 180^{0}

y = 180^{0} – 83

y = 97^{0}

x+107^{0} = 180^{0}

x= 180^{0 }– 19

**x = 63 ^{0}**

**2. THEOREM**:

Any inscribed angle in a semicircle is a right angle.

**Given **

AB – Is a diameter at a circle.

O – Is the center.

C – Is any point on the Circumference

Required to prove that ACB = 90^{0}

2 ACB= AÔB (Angle at the centre)

But,

AÔB = 180^{o} (Straight angle)

ACB = 90^{o} Hence proved.

Example:

Calculate the Value of x

Soln:

PQR = 90^{o} (Angle in a semi Circle)

Then 90^{o} + 30^{o} + x = 180^{o }

(Angles sum in A)

x = 180^{o} – 120^{o}

x = 60^{o}

An exterior angle is an angle formed outside a cyclic quadrilateral. An internal angle is formed inside the cyclic quadrilateral when you provide line from this angle you will form an angle which is outside called exterior angle.

**THEOREM: 3**

Exterior angle of a cyclic quadrilateral is equal to the inside opposite angle.

Aim: to prove that PSR = PQT

Proof: let angle PQR = a

Angle PRS = b

Angle PQT = y

(i) PQR + PSR = 180^{o} (Because it is opposite angle of cyclic quadrilateral).

Thus a+b = 180^{o}

(ii) Therefore when you equate them since both are 180^{o}

(iii) PQR + RQT = 180^{o} (Because they are adjacent angles on straight line)

Thus a + y = 180^{o}

Therefore when you equate them since both are 180^{o} you will have:

a + b = a + y

B = y

But b = PSR

And y = PSR

PSR = PQT

Proved.

**THEOREM 4**:Angles in the same segment are equal.

**THEOREM 5**: Angles in the semi circle are right angled triangle.

Aim: To prove that PxQ = PyQ

Construction: join OP and OQ

Let PÔQ = P

PVQ = q

PYR = r

(i) PÔQ = 2p x Q and

P = 2q (Because angle at the centre is twice the angle at the circumference)

(ii) PÔQ = 2PyQ (Because angle at the centre is twice the angle at the circumference)

P = 2r

x = P = 2q, P = 2r

= 2q = 2r

q = r

** TO ANSWER THE QUESTIONS:**

1. Find the value of X. if O is the centre of Circle

Soln:

Angle at the centre is twice the angle at the circumference.

Since the circle at the centre is 360^{o}

3 x + 2x = 360^{o}

5x = 360^{o}

= 72^{o}

(b) Angle at the centre is twice the angle of the circumference.

Opposite angle at a cyclic quadrilateral are supplementary (add up to 180^{o})

x + 3x = 180^{o}

4x = 180

x=45

Soln:

Angle at the same segment are equal.

Angle at the centre is twice the angle at the circumference.

2 x + 260^{o} = 360^{o}

Since the circle at the centre is 360^{o}

2x + 260^{o} = 360^{o}

2x = 360^{0}-260^{0}

2x=100^{0}

x = 50^{o}

Soln:

Angle at the centre is twice the angle at the circumference.

150^{o} x 2 = 300^{o}

Since the circle at the centre is 360^{o}

x + 300^{o} = 360^{o}

x = 360^{o} – 300

x = 60^{o}

Soln:

Angle at the centre is twice the angle at the circumference.

40^{o} x 2 = 80^{o}

Since the circle at the centre is 360^{o}

x + 80^{o} = 360^{o}

x = 360^{o} – 80^{o}

x = 280^{o}

(f) Find the value of angles marked x, y and z

Soln:

Angle at the centre is twice the angle at the circumference.

X x 2 = 224^{o}

X = 112^{o}

Since the circle at the centre is 360^{o}

y + 224^{o} = 360^{o}

y = 360^{o} – 224^{o}

y = 136^{o}

Opposite angle of a cyclic quadrilateral are supplementary (add up to 180^{o})

Z + 112^{o} = 180^{o}

Z = 180^{o} – 112

z = 68^{o}

(g) Find ‘y’

Soln:

Angle at the centre is twice the angle at the circumference.

y x 2 = 2y

Since the circle at the centre is 360^{o}

4y + 2y = 360^{o}

y = 60^{o}

** Class activity**

1. MN is a diameter of a circle and L is a point on the circle. If MNL = 135^{o},

Find NML

Soln:

Any inscribed angle in a semicircle is a right angled triangle.

let NML=X

X + 35^{o} + 90^{o} = 180^{o}

X + 125^{o} = 180^{o}

X = 180^{o} – 125^{o}

NML = 55^{o}

2. AB is a diameter of a circle radius 10 cm and e is a point on the circumference.

CB = 12, find CA (Remember Pythagoras theorem)

Soln:

By using Pythagoras theorem

20^{2} = CA^{2}+ 12^{2}

400 = CA^{2}+ 144

CA^{2}= 400 – 144

CA^{2 } = 256

CA= 16cm

3. AB is a diameter of a circle radius 40cm and C is point on the circumference.

If CBA = 62^{O}, then find CÂB

Soln:

Let X = CÂB

X + 62^{o} + 90^{o} = 180^{o}

X + 152^{o} = 180^{o}

X = 180^{o} – 152^{o}

CÂB = 28^{o}

**THE CHORD PROPERTIES OF A CIRCLE**

The chord of a circle is the line segment whose end point are on the circle. A chord which passes through the centre of a circle is called a diameter. It is very important for you to know what a chord is and how to identify the chord properties of a circle because it will summarize you with this unit.

Therefore in this section you are going to study about the chord itself and the **chord properties** of a circle.

You are also going to study how to develop theorem which relate to these properties at chord. At the end of the section you will be able to identify the chord, prove the theorem of the chord. Properties in a circle and then apply these theorems on solving related problems in order to identify the properties of the chord properties it easier if you draw a circle with centre O.

You can see that O is a centre of the OM is the radius of the circle and PQ is chord of the circle.

Therefore you will discover that.

(a) The centre of the circle lies on the perpendicular bisector of the chord.

(b) The perpendicular from the centre of the circle to the chord

(c) The line joining the centre of the circle to the midpoint of the chord.

Then from the information above you can develop the theorem which can be written as;

**THEOREM**

The perpendicular bisector of a chord passes through the centre of the circle.

AIM: To prove that OMP = OMQ = 90^{o}

Construction: Join OP, OQ and OM

Proof:

(i) OP = OQ (Radii)

(ii) PN = QM (M is midpoint given)

(iii) ON = OM (common)

(iv) OPM = OQM (Bisected angles)

The corresponding angle are congruent and hence OMP = OMQ = 90^{o} proved.

**THEOREM:** Parallel chords intercept congruence arc.

Aim: To prove that arc PR≡Are

Proof:

Arc AQ≡ Arc AP (AOB is diameter)

Arc AS≡ Arc AR (AOB is a diameter)

Arc PR ≡Arc AR – Arc AP and also

Arc QS≡ Arc AS – Arc AQ

By step (i) up to step (iii) above you can conclude that Arc PR≡ Arc QS proved

**Class Activity:**

Two chords, AB and CD of the circle whose radius is 13cm are equal and parallel.

If each is 12cm long, find the distance between them.

Soln: – By using Pythagoras theorem

AD^{2}= AB^{2 }+DB^{2 }

26^{2} =12^{2} + DB^{2}

676 = 144 + DB^{2}

DB^{2}= 676 – 144

DB^{2} = 532

Square root both sides

2.A chord of length 32cm is at a distance of 12cm from the centre of a circle.

Find the radius of a circle.

Soln:

By using Pythagoras theorem

AO^{2 } = 12^{2}+ 16^{2}

= 144 + 256

= 400

Apply square root both sides

3. A distance of a chord PQ from the centre of a circle is 5cm.

If the radius of the circle is 13cm. Find the length of PQ

Soln:

By using Pythagoras theorem.

13^{2} = 5^{2} + PQ^{2}

169 = 25 + PQ^{2}

PQ^{2}= 169 – 25

PQ^{2}= 144

Apply square root both sides

PQ = 12cm

Since OS is perpendicular to PQ

Therefore PS = SQ

PQ = 12cm + 12cm

= 24cm.

4. The chord AB of a circle with centre O radius 3cm long.

Find the distance of AB from O. give your answer in cm form

Soln:

By using Pythagoras theorem

a^{ 2} +b^{2} =c^{2 }

15+b^{2} =9 x 3

b^{2} =27-15

Apply square root both sides

∴

5.Two chords AB and CD of a circle with centre O. if AB = 10cm,

CD = 6cm, AO = 7cm. Find the distance between two chords

Triangle (i)

By Pythagoras theorem

a^{2} x 6 = C^{2}

5^{2} + b^{2} = 7^{2}

25 + b^{2} = 49

b^{2} = 49 – 25

b^{2} = 24

triangle (ii) by Pythagoras theorem

a^{2} + b^{2} = c^{2}

3^{2} + a^{2} = 7^{2}

9 +a^{2} = 49

a^{2} = 49 – 9

a^{2} = 40

** Class Activity:**

XY and PQ are parallel chords in a circle of centre O and radius 5cm.

If XY = 8cm and PQ = 4cm, find the distance between two chords.

Soln:

1^{st} triangle

By using Pythagoras theorem.

C^{2} = a^{2} + b^{2}

5^{2} = 4^{2} + b^{2}

25 = 16 + b^{2}

b^{2} = 25 – 16

b^{2} = 9cm

square root both sides

b= 3cm

2^{nd} triangle

By using Pythagoras theorem.

C^{2} = a^{2} + b^{2}

5^{2} = 2^{2} + b^{2}

25 = 4 + b^{2}

b^{2} = 25 – 4

** Example:**

1. AB is a chord of circle with centre X. the midpoint of AB is m.

If XAB find MXA

XAM + MXA = 180^{O}

52^{o} + 90^{o} + MXA = 180^{o}

MXA = 180^{o – }142º

= 38^{o}

2. AB is a chord in a circle with centre C. the length of AB is 8cm and the

Radius of the circle is 5cm. find,

The shortest distance of AB from C

ACB

By applying Pythagoras theorem.

4^{2} +mc^{2} = 5^{2}

Mc^{2} = 25 – 16

Mc^{2}= 16

Square root both sides

Mc = 4m

Sin ACM = 0.8000

ACM = Sin^{-1} ((0.8)

= 53^{o} X 2

= 106^{o}

A chord AB has length 12m. it is 7m from the centre of the circle.

Find the (a) length of AC

(b)ACB

Soln:

By using Pythagoras theorem.

Using:

Tan ACM = 0.8571

ACM = Tan^{-1}

= 40^{o}

ACB =40^{0} x 2

= 80^{o}

**EXERCISE**

1. M is the Centre at the chord AB at a Circle with centre X if

2. M is the centre at chord PQ at a circle with Centre O. if < PQO = 43^{O}. Find

3. A circle has radius 13cm and centre X. a chord AB has length 24cm. find:-

(a) T distance of the chord from the Centre

(b) < AXB

By Pythagoras theorem

(AM)^{2} + (MX)^{2} = (AX)^{2}

12^{2} + (MX)^{2} = 132

144 + (MX)^{2} = 169

(MX)= 169 – 144

(MX)^{2 }= 25

Square root both sides

(MX)^{2 }= 25

MX = 5cm

The distance from the centre to chord is 5cm.

**QUESTIONS:**

1. Let A be the Centre at a chord PQ at a circle with Centre O. If < PQO = 43^{o}, find < POQ

Soln

1. Q is the centre at a Circle and AB is a Chord

(a) The length at AB

(b) The distance at A from C

Soln:

3. PQ is a Chord in a Circle with centre R. PQ = 14cm and the distance at R from PQ is 9cm, find:-

(a) The radius at a circle

(b) <PRQ

Soln:

Let RQ = Radius

By using Pythagoras theorem

Using, SO TO CA

H A H

**TANGENT PROPERTIES**

A tangent to a circle touches it at exactly one point

**THEOREM:**

A tangent to a circle the line perpendicular to the radius at the point of contact.

TAP is a line perpendicular to the radius CA show that TAP is a tangent as follows:-

If b is another point on TAP then CB is the hypotenuse at A CAB and hence CB is longer than CA it follows that B lies outside the Circle. Hence TAP needs the Circle only at A. TAP is a tangent to the Circle.

Hence a tangent is perpendicular to the radius.

**Examples:**

1. TA is a tangent to the Circle with centre C. If

Line AT⊥AC = 90°

<CAT+<ACT +<ATC=180°

90° + 49° + <ATC =180°

139° + <ATC =180°

<ATC=180° – 139°

∴ <ATC = 41°

2. A point T is 8cm from the centre C of a circle of radius 5cm. Find

(a) The length of the tangent from T to the circle

(b) The angle between the tangent and TC

By using Pythagoras theorem

(AC)^{2} + (AT)^{2} = (TC)^{2}

5^{2} + (AT)^{2} = 8^{2}

25 + (AT)^{2} = 64

AT^{2} = 64 – 25

(AT)^{2} = 39

AT = 6.24

=39º

**QUESTIONS:**

1. TA is a tangent to the circle at A. the centre is C. if <CTA=32º

(a) Find <ACT

(b) If TC = 8cm, Find AT and radius of the circle

**Solution: (a)**

Line AT ⊥ AC = 90^{0}

<ACT +=180^{0}

<ACT +90^{0} + 32º =180^{0}

<ACT= 180^{0}– 122^{0
} <ACT= 58^{0}

**Solution: (b)**

= 8 (0.5299)

AT = 4.2392cm

2.TA is a tangent to a circle at A. The centre at the circle C if angle ACT = 73°and radius of the circle is 2m. find:-

(a) <ATC

(b) TA and TC

Line TA AC = 90^{0
<ATC+<ACT +<CAT=180°
90° + 73° + <ATC =180°
}

<ATC + 163^{0} = 180^{0}

<ATC= 180^{0} – 163^{0
<ATC =17º
}

**Solution: (b)
(i)
**

Opp = 2 Tan 73^{0}

=2(3.2709)

TA = 6.5418m

(ii) By using Pythagoras theorem

TA = 6.5418m

≈7m

C^{2} = a^{2} + b^{2}

= 2^{2} + 7^{2}

= 4+49

c^{2}= 53

Square root both sides

C^{2} = 53

3. A point T is 6m form the center C of a circle radius 3cm. Find:-

(a)The length of Tangent from T to the circle

(b)The angle between the tangent and TC

**Solution: (a)**

Using Pythagoras theorem

A^{2} + b^{2} =C^{2}

3^{2} + b^{2} = 6^{2}

9 + b^{2 }= 36

b^{2} = 36 – 9

b^{2} = 27

Square root both sides

b^{2} = 27

**Solution: (b)
**

Cos = 0.5

= Cos -1 (0.5)

= 30^{0}

Angel between tangent and TC is 30^{0}

**Class Activity**

1.A point, 10m from the center X of circle at radius 6m. A tangent is drawn from P to the circle touching at A. Find the length of the tangent from P to the circle.

By using Pythagoras theorem

C^{2} = a^{2} + b^{2}

10^{2} = 6^{2} + b^{2}

100 = 36 + b^{2}

b^{2} = 100 – 36

b^{2} = 64

Square root both sides

b^{2} = 64

b = 8m

∴The length of the tangent from P to the circle is 8m

2.A tangent is drawn from T to a circle of radius 8cm. The length of the tangent is 4cm. Find,

(a) The distance of T from the Centre C of the circle

(b) The angle between TC and the tangent

(a) By using Pythagoras theorem

C^{2} = a^{2} + b^{2}

= 8^{2} + 4^{2}

= 64 + 16

= 80

Square root both sides

C^{2} = 80

θ = tan^{-1} (2)

= 6^{0}

the angle between TC and tangent is 63º

**TANGENT FROM A POINT**

Suppose T is outside a circle there are two tangent from T to the circle and they are equal in length

**Proof:**

Consider the angle TCA and TCB

CA = CB (Both are radii)

TC = TC (Common)

<TAC=<TBC

**CHORD AND TANGENT**

Suppose the chord CD gets shorter and shorter is that C and D approach a common point E then the chord CD becomes the tangent at E, by interesting chord theorem.

XA x XB = AC x XD

XA x XB = XE x XE

XA x XB = (XE)^{2}

**Example**

1.TX is a tangent to a circle. The line TAB cuts the circle at A and B with TA = 3cm and AB = 9cm. Find TX

**Solution:**

(TA) (TB) = TX x TX

(TA) (TB) = (TX)^{2}

Since, TB = (TA) + (AB)

= 3cm + 9cm

= 12cm

From the theorem

(TA) (TB) = (TX)^{2}

3cm x 12cm = (TX)^{2}

Square root both sides

36cm^{2} = (TX)^{2}

TX = ± 6cm

Since there is no -ve dimension therefore TX is 6cm

**More Examples:**

Find the length of unknown in the diagram.

Solution:

(CB) (CA) = (DC) (DC)

(CB) (CA) = (DC)^{2}

Since CA = (CB) + (AB)

= 2m + a

From the theorem.

(CB) (CA) = (DC)2

(2m) (2m +a) = (4m)^{2}

4m^{2} + 2ma = 16m^{2}

2ma = 16m^{2} – 4m^{2}

2ma = 12m^{2
}a = 6m

Solution:

(CB) (CA) = (CD)^{2}

Since CA = (CB) + (AB)

= 9m + 7m

= 16

From the theorem

(CB) (CA) = (CD)^{2}

(9m) (16m) = (b)^{2}

Square root both sides

144m^{2} = b^{2}

b = ±12m

Since there is no -ve dimension **b= 12m**

2.TC is a tangent to a circle and Tab cuts at AB and B. if TA = 2cm and TB = 8cm, find TC

**Soln:**

From the theorem

(TA) (TB) = (TC)^{2}

2cm x 8cm = (TC)^{2}

16cm^{2} = (TC)^{2}

Square root both sides

16cm^{2} = (TC)^{2}

TC = 4cm

3. TX is a tangent to a circle and TYZ cuts the circle and Y and Z. if TX= 10m and TY = 4m. Find TZ

Solution:

(TY) (TZ) = (TX)^{2}

Let ZY = y

Since TZ = TY + ZY

= 4m + y

**From the theorem**

(TY) (TZ) = (TX)^{2}

(4m) (4m + y) = (10m)^{2}

16m^{2} + 4my = 100m^{2}

4my = 100m^{2} – 16m^{2}

4my = 84m^{2}

Y = 21m

Since ZY = y

TZ = TY + ZY

= 4m + 21m

= 25m

**Class Activity**

1.TA is a tangent to a circle and TBC meets the circle at B and C. TA = (9cm and BC = 24cm). Find TB

Solution:

(TB) (TC) = (TA)^{2}

Let (TB) = y

Since TC = TB + CB

= y + 24cm

From the theorem

(TB) (TC) = (TA)^{2}

(y) (y+ 24cm) = (9cm)^{2}

Y^{2} + 24cmy = 81cm^{2}

Y^{2} + 24cmy – 81cm^{2} = 0

By using General formula

Where a = 1, b = 24, c = 81

Since there is no negative dimension, the length at TB is 3cm

2.TX is a tangent to a circle and TPQ meets the circle at P and Q. TX = 12cm and PQ = 7cm, find TP

Solution:

(TP) (TQ) = (TX)^{2}

Let (TP) = Z

Since TQ = TP + QP

= Z + 7cm

From the theorem

(TP) (TQ) = (TX)^{2}

(z) (z +7cm) = (12cm)^{2}

z^{2} + 7z = 144

z^{2} + 7z – 144 = 0

By using the general formula

Where a = 1, b = 7 and c = -144

Since there is no negative dimension, the length at TP is 9cm

3.AB is a chord at a circle at length 5cm. C is another point on the circle. AB extended on the circle meets the tangents at C and T. if the TC = 6cm, find the possible value of TB.

**Solution:**

(TB) (TA) = (TC)^{2}

Let

(TB) = x

Since TA = TB + AB

= x + 5cm

**From the theorem**

(TB) (TA) = (TC)^{2}

(x) (x + 5) = (6cm)^{2}

X^{2} = 5x = 36

X^{2} + 5x – 36 = 0

**By completing the square**

X^{2} + 5x – 36 = 0

X^{2} + 5x – 36 = 0

X^{2} + 5x = 36

Add (½ b)^{ 2} both sides

X^{2} + 5x + (½ x 5) = 36 + (½ x 5)^{2}

X^{2 }+ (5/2)^{2} = 36 + 25/4

(x + 5/2)^{2} = 169/4

**Square root both sides**

(x + 5/2)^{2} = ±169/4

X + 5/2 = ± 13/2

X = -5/2 ± 13/2

= -5/2 + 13/2

= 4cm

**Or**

X = -5/2 – 13/2

= -9cm

Since there is no negative dimension, the length of TB is 4cm.

4. XY is a chord of a circle at length 2cm. z is another point on the circle. XY extended meets the length at z at T. if TX = 18cm, find the possible value of TZ

Solution:

(TY) (TX) = (TZ)^{2}

Since TX = TY + XY

18cm + 2cm= 20cm

From the theorem

(TY) (TX) = (TZ)^{2}

(18cm) (20cm) = (TZ)^{2}

360cm^{2} = (TZ)^{2} square root both sides

(TZ)^{ 2} = 360cm^{2}

**ALTERNATE SEGMENT THEOREM**

AT is a tangent to the circle and AB is a chord. The alternate segment theorem state that:-

THEOREM: The angle between the chord and tangent is equal to the angle in the alternate (others)

Segment i.e. < TAB = <ACB

**Proof:**

< TAB = <ACB

**Aim:** Is proving that <ACB=<BAT

Let ACB = X and Centre of a Circle to be

‘O’ AOB = 2x (< at the centre is twice the angle at the circumference).

AC,BC, and AB or chords in the Circle and AO = OB

∴ΔAOB is isosceles triangle since AO and OB are equal

<OAB=1/2(180º – 2x)

= 90^{0} – x

<BAT+ <OAB =<OAT

<BAT=<OAT – <OAB

=90° – (90° – x)

=90^{0}-90^{0} +x

=0+x

∴<BAT=x

hence proved

∴<ACB=<BAT

=x

**QUESTIONS:**

1.ABCD is a cyclic quadrilateral TA is the tangent to the Circle at A. if <TAC = 73º, find <ABC

Solution:

TAC = 73^{0}

Angle at the same segment equal

TAC = ADC

<ADC + 73^{0} = 180^{0}

Take out 73^{0} both sides

<ABC = 180^{0} – 73^{0}

<ABC=107º

2. PQRS is a cyclic quadrilateral. ST is the tangent to the Circle at S. if <QRS = 132º.Find <RST

Soln:

<QRS +<RST =180º

132^{0} +132^{0}=180º

Take out132^{0} both sides

<RST =180º-132^{0}

=48º

3. C is the centre of the Circle. If<BAT=39º. find <ACB.

**Soln:**

Angle at the Centre is twice the angle at circumstances

<ACB = <BAT

39º=<BAT

<ACB x 2=**<** ABC

< ABC=39º x 2

=78º

4.X is the Centre of the Circle if<CXD =88º, find <TCD

Soln:

<DXC=<DCT/2

Angle at the Centre is twice the angle at the circumstances.

∴<TCD=88º/2

= 44º

**THE EARTH AS A SPHERE**

**SPHERE**

Is a set of a point which equidistance (equal distance) from the fixed point called the centre of the Sphere.

–The distance from the centre of the sphere to any point at the circumference of the sphere called Radius at the earth which is approximately as 6370km.

–The surface of the earth is not exactly spherical because it is flattened in its northern and southern pole or we say. The earth is not perfect sphere, as it is slightly flatter at the north and southern pole than at the equator. But for most purpose we assume that it is a sphere

**THE EARTH AS PERFECT SPHERE**

We consider the earth to be a perfect sphere of radius 6370km at approximately 6400km.

–Whereby O is the center of the earth.

–R is the radius of the earth

–The earth rotates once a day about a line called polar axis (earth axis)

–This axis basses through the center and joins the northern and southern poles

The equator has 0^{0}

**DEFINITIONS OF TERMS**

I. **GREAT CIRCLE (EQUATOR)**

Is the line which drawn from west to east with 0^{0}. Or

Is an imaginary line which divides the earth surface into two equal parts: Southern part and Northern part through the Centre of the earth called hemisphere.

The equator is the only Great Circle perpendicular to the earth axis

–Earth point on the earth’s surface is said to be either in northern hemisphere or southern hemisphere.

II. **SMALL CIRCLE (LATITUDE)**

Is the line drawn from West to east and measure in degree from the centre at the Earth (Equator) Northward or Southward.

– Latitude range from 0^{0}N or 90^{0}S

– The radius of parallel at latitudes becomes smaller as one moves towards the southern or northern pole

Example of the line of latitude that should be drawn from west to east .

– The equator is the standard zero latitude from which other latitude are measured

– Any other line of latitude is named by the longitude basses through when rotating from the Equator to the line of latitude this angle is either north or south of the equator.

When naming a latitude it is essential to say whether it is north or south of the Equator.

III. **MERIDIAN (LONGTUDE)**

– These are lines drawn from north to south measured degree from the prime meridian westward or eastward.

– These circles are not parallel as they meet at the poles. These circles have radius equal to that of the earth and they are called great circle.

– Longitudes are also called meridian.

**Diagram:**

– In order to name lines at longitudes it has been necessary to choose a standard zero called the prime meridian.

This is a line at longitude which passes through Greenwich, London.

IV.**THE GREENWICH MERIDIAN**

–Greenwich meridian: Is a longitude whose degree measure is zero (0^{0})

–Greenwich meridian is also considered as prime meridian

–Greenwich meridian is standard longitude in which other meridian are measured in degree from East for West.

– The Greenwich divides the earth’s into two parts eastern part and western part

– Each point on the earth surface is said to be either on the eastern part or western part.

Diagram:

– The longitude of a place varies from 0^{0} along the Greenwich meridian to 180^{0}E or 180^{0}W. The Radius of all longitude are equal to that of the earth.

**LOCATION OF POINTS ON THE EARTH SURFACE**

1.O is the centre of the earth. The latitude of P is 50^{0}N and of Q is 40^{0}S

**Solution:**

Since the given point has been allocated into different hemisphere therefore the angle subtended by an arc PQ = 50^{0} + 40^{0}= 90^{0}

2. Two towns are on the same circle of longitude. One town is 20^{0}N and other is 30^{0}S. What is the angle subtended by an arc of these two angles.

Solution:

Since the two towns has been found into different hemisphere the angle subtended by an arc of the two angle are

20^{0} + 30^{0} = 50^{0}

3. Two towns C and D line on the equator. The longitude of C is 70^{0}E and for D is 30º

E.

What is the angle subtended by an arc.

Solution:

Since C and D are in the same hemisphere the angle subtended by an arc CD = 70^{0} – 30^{0}

CD= 40^{0}

4. Two towns A and B are on the equator.

The longitude of A is 35^{0}E and the B = 72^{0}W. Find the angle subtended by an arc AB

Since the point given has been found in the different hemisphere the angle subtended by an arc AB = 72^{0} + 35^{0}

= 107^{0}

**HOME WORK**

1.Given that Morogoro is (7^{0}S, 38^{0}E) and Moscow is (56^{0}N, 38^{0}E). Find the angle subtended by the area which connect the two places at the centre of earth.

**Solution:**

Since the given places have been alocated in the different hemisphere therefore subtended by an arc = 56^{0} + 7^{0}

= 63^{0}

2. What is the difference in longitude between Brazivile Congo (4^{0}S, 15^{0}E) and Mombasa Kenya( 4^{0}S, 40^{0}E)

**Solution:**

Since the given places have been allocated in the same hemisphere therefore the different between the two places = 40^{0} – 15^{0}

= 25^{0}

**DISTANCE BETWEEN TWO PLACES MEASURED ALONG GREAT CIRCLE**

Note: Great circles means either the equator or lines of longtudes.

Consider two points P and Q both found on the equator.

Diagram:

O is the centre of the earth OP = OQ = Radius of the earth angle POQ is the central angle line PO = ‘l” is the length at arc PQ on the equator.θ is the difference in longitudes between points P and point Q.

Remember, If P and Q are on the same hemisphere. θ will be found by subtracting their respective longitudes and if P and Q are in different hemisphere, the value ofθ will be obtained by taking their sum of the respective longtude.

θ= l

360^{0} = 2πR

Whereby, π= 3.14

R = 6370km

Example:

Three points A(0^{0},14^{0}W), B(0^{0},25^{0}W) and C(0^{0}, 46^{0}E) are on the Earth’s surface.

Calculate the length of the equator

(a)AB (b) AC (c) BC

The distance of points A and B is 1223km

(b)

The distance of point AC is 6668km

(c)

**The distance of point BC is 7891**

**HOME WORK**

1.Three points are such that A(43^{0}N, 10^{0}E), B(16^{0}N, 10^{0}E) and C(28^{0}S, 10^{0}E). calculates the lengths at the following arcs measured along the longitudes.

(a) AB (b) BC (c) AC

Solutions:

The distance at point AB is 3000km.

(b)

The length at point BC is 6156km.

(c)

The distance of point AC is 7,891km.

**Home Work**

1.Two towns R and Q are 2813km apart R being direction of North of Q. If the latitude Q is 50S, find the Latitude of R.

Solution:-

Data

Distance RQ = 2813km

R = Required

Q = 5^{0}S

**
**Since R is due to North of Q,the latitude of R is 20.33º

**Class Work**

1.Calculate the distance between Tanga (5^{0}, 39^{0}E) and Ruvuma (12ºS, 39°E) in;

(a) Nautical mile

(b) Kilometre

(a)Solution:

(α ± β) 60^{0} = Nm

(12^{0} – 5^{0}) 60^{0 }= Nm

7^{0} x 60^{0} = Nm

(b) 1Nm = 1.852km

420.1Nm =?

The distance between Tanga and Ruvuma is

(a) 420. 1Nm

(b) 778km

1.Find the distance between point x and y given that X(34^{0}N, 124^{0}E) and Y(41^{0}N,124^{0}E)

(a)In nearest Nautical mile

(b)In nearest Kilometers

Solutions:

( β) 60^{0} = Nm

(41^{0} – 34^{0}) 60^{0} = Nm

7^{0} x 60^{0} = Nm

= 420.1Nm

The distance between point X and Y is

(a)420 Nm

(b)793Km

**Home Work**

1.An air Craft took a height from town A(4^{0}N, 12^{0}E) moving southwards along a great circle for a distance of 2437Km. write the position of the town it landed.

Solution:

Distance = 2437Km

A = 4^{0}N, 12^{0}E

B = Required (5º 12^{0}E)

4° + β = 21.93°

β = 21.93^{°} – 4^{°
=17.930
}

The position of town of Landed is 17.93^{0}

**DISTANCE BETWEEN TWO POINTS ALONG THE PARALLEL OF LATITUDE.**

Except the equator, other parallels of latitudes are small circles. Simply because their radii are less than the radius of the earth

Consider the point P and Q both found on the same parallel of latitude, let say α^{0}N and Q is the different in longitudes between P and Q.

R is the radius parallel latitude of α^{0}N

Diagram

Length at arc PQ = θ

Circumference 0f small circle 360^{0
}

But; r is the radius of small circle.

But PQ is the distance between two points but θ = α ± β

Example

1.Calculate the distance between P(50^{0}N, 12^{0}W) and Q(50^{0}N, 26^{0}E)

Solution:

**Home Work**

1. Find the distance between point A(58^{0}N, 23^{0}E) and (58^{0}N, 40^{0}W)

Solution

The distance between point A and B is 370Km

**CLASS ACTIVITY**

1. Calculate the distance from town P and Q along the parallel of latitude.

If P (23^{0}N,10^{0}E) and Q (23^{0}N, 54^{0}E)

Solution:

4.501 x 1000 =4501 x 1

= 4501km

The distance from P to Q is 4501 Km

2. Two Towns both on latitude 45^{0}S differ in longitude by 50^{0}. Calculate the distance between two towns measured along the parallel of latitude.

Solution:

^{}

3.929 x 1000 = 3929 x 1

=3929Km

The distance between two towns is 3929km.

**Example;**

A plane flying at 595km/hour leaves Dar es Salaam (7^{0}S, 39^{0}E) at 8:00 am. When will it arrive at Addis Ababa at (9^{0}N, 39^{0}E)

Solution:

1.778 x 1000

1778km

The distance from Dar to Addis Ababa (Ethiopia) is 1778km

= 2:54

Since it spent 2:54 and left at Dar around 8:00 am now will reach Addis Ababa at 10:54am

**CLASS ACTIVITY**

1.An aero plane flies from Tabora (50S, 330E) to Tanga (5ºS, 39ºE) at 332 km/hour along parallel of latitude. If it leaves at Tabora at 3:00 pm. Find the arrival time at Tanga airport.

^{Solution.}

664.3km

The distance between Tabora and Tanga is 664km.

= 2 hours

Time arrival time at Tanga air port is 3:00pm + 2:00pm = 5:00pm

**Home Work**

1.A ship is Steaming in a eastern direction from town A to town B. if the position A is (32^{0}N, 136^{0}W) and B is (32^{0}N, 138^{0}W). What is the speed of sheep

if it takes 3hours from town A to town B.

Solution:

**
**2.73 X 1000=2073Km

The distance from town A to B is 2073km.

The speed of sheep is 691km/hour.

In solving problems involving speed at ships, a term known as knot is usually used. By definition a speed of one nautical mile per hour is called knot

Therefore

1knote = 1Nm/hour = 1.852km/hour

Example: 1

When a ship is given 20knots is actually sailing at 20 nautical miles per hour or approximately 37 kilometer per hour.

Example: 2

A ship sails northwards to Tanga (5^{0}S,39^{0}E) at an average speed of 12 knots. If the ship starting points is Dar es Salaam (7^{0}S, 39^{0}E) at 12:00noon, when will it reach Tanga.

Solution: 1

How did 37km obtained.

1knote = 1.852km/hour

= 2 x 18.52km/hr

= 37.04km

= Approximately 37km

Solution: 2

^{}

1knot = 1.852km/hr

12knot= x

1knot x = 12knots x 1.852km/hr

X = 10.09

= 10:00

∴From 12:00 noon adding 10hours will sail at Tanga at 10:00pm.

**Home Work**

1. A ship is teaming at 15knots in western direction from Q to R. if the position of P is 40ºS, 178ºe and that of Q is 40ºS, 172ºE, how long will the journey take?

Solution:

510.7 km is the distance from point Q to R.

= 18:12

The journey took = 18 hours and 12 minutes

**Class Activity**

1. A speed boat traveling from Zanzibar (6^{0}S, 45^{0}E) to Mwanza (9^{0}S, 45^{0}E) using 30knots left Zanzibar at 11:30am at what time did it reach at Mwanza?

Solution:

The distance from Zanzibar to Mtwara is 333.4km

= 5:54pm

Since it spent 5:54pm and left at Zanzibar at 11:30am. Now it will reach at Mtwara 5:24pm.

2. Find the time taken for a ship to sail from town P(80^{0}N, 60^{0}W) to town Q(60^{0}S, 60^{0}W) in 70knots

Solution:

=129.64km/hr

≈130km/hr

= 119:36

The ship will take 119hours and 36 minutes from town P to town Q

**Class Activity.**

1. A ship sails from A (0^{0},20^{0}W) to B (10^{0}N, 20^{0}W) at 16 knots. If it leaves A at 8:00am on Tuesday when will it reach B?

Solution:

1knot = 1.852km/hour

16knots =?

= 16 x 1.852km/hour

= 29.632km/hour

30km = 1hour

= 37hours

37 – 24hour = 13hours

8:00 + 24hr = 8:00am

8:00am + 13hours = 9:00am

The ship will reach town B at 9:00 am on Wednesday

**Home Work**

A ship sails from point A (10^{0}S, 30^{0}W) to B(10^{0}N, 30^{0}W) at 20 knots. If it leaves point A at 12:00 midnight on Monday when will arrive at B?

Solution:

1kont = 1.852km/hour

20knot = x?

knot x = 20knot x 1.852km/hour

≈37km/1hour

2222km = x?

37km x = 2222km x 1hour

=60hours

Since 12:00pm on Monday the ship has spend 2 and ½ days where by it will arrive at B at 12:00 afternoon on Thursday.