# Circles, Chords and Tangents - Mathematics Form 3 Notes

## Length of an Arc

• The Arc length as shown in the figure above is given by;

θ   x 2πr. where r is the radius of the circle
360

Example

Find the length of an arc subtended by an angle of 2500 at the centre of the circle of radius 14 cm.

Solution

Length of an arc = θ  x 2πr
360
= 250 x 2 x 22 x 14 = 61.11 cm
360        7

Example

The length of an arc of a circle is 11.0 cm.Find the radius of the circle if an arc subtended an angle of 900 at the centre.

Solution

Arc length = θ  x 2πr but θ = 900
360

Therefore 11 = 90 x 2 x 22 x r
360         7
r = 7.0 cm

Example

Find the angle subtended at the centre of a circle by an arc of 20 cm, if the circumference of the circle is 60 cm.

Solution

= θ   x 2πr = 20
360
But 2πr = 60 cm
Therefore,  θ  x 60 = 20
360
θ = 20 x 360
60
θ = 120
0

## Chords

• Chord of a circle: A line segment which joins two points on a circle.
• Diameter: a chord which passes through the center of the circle.
• Radius: the distance from the center of the circle to the circumference of the circle

### Perpendicular Bisector of a Chord

• A perpendicular drawn from the centre of the circle to a chord bisects the chord.

• Note;
• Perperndicular drawn from the centre of the circle to chord bisects the cord (divides it into two equal parts)
• A straight line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord.

Example

The radius of a circle centre O is 13 cm.Find the perpendicular distance from O to the chord, if AB is 24 cm.

Solution

OC bisects chord AB at C
Therefore, AC =12 cm
In ∆AOC ,OC2 = AO2  AC2
= 132 - 122 = 25
Therefore, OM = √25 = 5 cm

### Parallel Chords

• Any chord passing through the midpoints of all parallel chords of a circle is a diameter

Example

In the figure below CD and AB are parallel chords of a circle and 2 cm apart. If CD = 8 cm and AB= 10 cm, find the radius of the circle

Solution

• Draw the perpendicular bisector of the chords to cut them at K and L .
• Join OD and OC
• In triangle ODL,
• DL = 4 cm and KC =5 cm
• Let OK = x cm
• Therefore ( x + 2)2 + 42 = r2

In triangle OCK;

• x2 + 52 =r2
• Therefore ( x + 2)2 + 42 = x2 + 52
• x2 + 4x + 20) + 42 = x2 + 52
• 4x + 20 = 25
• 4x = 5
• x =

Using the equation x2 + 52 = r2
r2 = (5/4)2 52
= 25/16 + 25
425/16
r =√(425/16)
= 5.154 cm

Example

A hemispherical pot is used for a hanging basket. The width of the surface of the soil is 30cm. The maximum depth of the soil is 10 cm. Find the radius of the pot

Solution

c2 = a2 + b2
r2 = 152 + (r − 10)2
r2 = 225 + r2 − 20r + 100
20r = 325
r = 325/20
r = 16.25 cm

### Intersecting Chords

In general DE = EB or DE ×EC = EB × AE
AE    EC

Example

In the example above AB and CD are two chords that intersect in a circle at Given that AE = 4 cm, CE =5 cm and DE = 3 cm, find AB.

Solution

Let EB = x cm
4
× x = 5 × 3
4x = 15
x = 3.75 cm
Since AB = AE + EB
AB = 4 + 3.75
= 7.75 cm

### Equal Chords.

• Angles subtended at the centre of a circle by equal chords are equals
• If chords are equal they are equidistant from the centre of the circle

### Secant

• A chord that is produced outside a circle is called a secant
BC = CD OR BC × CA = CD × EC
EC     CA

Example

Find the value of AT in the figure below. AR = 4 cm, RD = 5 cm and TC = 9 cm.

Solution

AC x AT = AO x AR
(x + 9) x = (5 + 4) 4
x
2 + 9x = 36
x
2 + 9x − 36 = 0
(x + 12) (x − 3) = 0
Therefore, x = − 12 or x = 3
x can only be a positive number not negative hence x = 3 cm

## Tangent and Secant

#### Tangent

• A line which touches a circle at exactly one point is called a tangent line and the point where it touches the circle is called the point of contact

#### Secant

• A line which intersects the circle in two distinct points is called a secant line (usually referred to as a secant).

In the figures below, A shows a secant while B shows a tangent

### Construction of a Tangent

• Draw a circle of any radius and centre O.
• Join O to any point P on the circumference
• Produce OP to a point P outside the circle
• Construct a perpendicular line SP through point P
• The line is a tangent to the circle at P as shown below.

Note;

• The radius and tangent are perpendicular at the point of contact.
• Through any point on a circle , only one tangent can be drawn
• A perpendicular to a tangent at the point of contact passes thought the centre of the circle.

Example

In the figure below PT = 15 cm and PO = 17 cm, calculate the length of PQ.

Solution

OT2 = OP2  PT2
= 172 - 152
= √64
OT = 8 cm

### Properties of Tangents to a Circle from an External Point

If two tangents are drawn to a circle from an external point

• They are equal
• They subtend equal angles at the centre
• The line joining the centre of the circle to the external point bisects the angle between the tangents

Example

The figure below represents a circle centre O and radius 5 cm. The tangents PT is 12 cm long. Find:

1. OP
2. Angle TPT'

Solution

1. Join O to P
OP2 = OC2 PC2 (pythagoras theorem)
OP
2 = 52 + 122
= 25 + 144
= √169
Therefore, OP = 13 cm
2. <TPT' = 2TPO( PO bisect < TPT')
<OTP = 900
∆TPO is a right angled at T
cos <TPO= 12/13 = 0.9231
Therefore, <TPO = 22.62 0
Hence <TPT1 = 22.62 0 ×2
= 45.240

### Tangent Problem

• The common-tangent problem is named for the single tangent segment that’s tangent to two circles.
• Your goal is to find the length of the tangent.
• These problems are a bit involved, but they should cause you little difficulty if you use the straightforward three-step solution method that follows.
• The following example involves a common external tangent (where the tangent lies on the same side of both circles).
• You might also see a common-tangent problem that involves a common internal tangent (where the tangent lies between the circles). No worries: The solution technique is the same for both.

Example

Given the radius of circle A is 4 cm and the radius of circle Z is 14 cm and the distance between the two circles is 8 cm. Find the length of the common tangent.

Here’s how to solve it:

1. Draw the segment connecting the centers of the two circles and draw the two radii to the points of tangency (if these segments haven’t already been drawn for you).
Draw line AZ and radii AB and ZY.
The following figure shows this step. Note that the given distance of 8 cm between the circles is the distance between the outsides of the circles along the segment that connects their centers.

2. From the center of the smaller circle, draw a segment parallel to the common tangent till it hits the radius of the larger circle (or the extension of the radius in a common-internal-tangent problem).
You end up with a right triangle and a rectangle; one of the rectangle’s sides is the common tangent. The above figure illustrates this step.

3. You now have a right triangle and a rectangle and can finish the problem with the Pythagorean Theorem and the simple fact that opposite sides of a rectangle are congruent.
The triangle’s hypotenuse is made up of the radius of circle A, the segment between the circles, and the radius of circle Z. Their lengths add up to 4 + 8 + 14 = 26. You can see that the width of the rectangle equals the radius of circle A, which is 4; because opposite sides of a rectangle are congruent, you can then tell that one of the triangle’s legs is the radius of circle Z minus 4, or 14 – 4 = 10.
You now know two sides of the triangle, and if you find the third side, that’ll give you the length of the common tangent.
You get the third side with the Pythagorean Theorem:
x2 + 10= 262
x2 + 100 = 676
x2 = 576
x=24
(Of course, if you recognize that the right triangle is in the 5 : 12 : 13 family, you can multiply 12 by 2 to get 24 instead of using the Pythagorean Theorem.)Because opposite sides of a rectangle are congruent, BY is also 24, and you’re done.
Now look back at the last figure and note where the right angles are and how the right triangle and the rectangle are situated; then make sure you heed the following tip and warning.
Note the location of the hypotenuse. In a common-tangent problem, the segment connecting the centers of the circles is always the hypotenuse of a right triangle. The common tangent is always the side of a rectangle, not a hypotenuse.
In a common-tangent problem, the segment connecting the centers of the circles is never one side of a right angle. Don’t make this common mistake.

### How to Construct a Common Exterior Tangent Line to Two Circles

• In this lesson you will learn how to construct a common exterior tangent line to two circles in a plane such that no one is located inside the other using a ruler and a compass.

Problem 1

For two given circles in a plane such that no one is located inside the other, to construct the common exterior tangent line using a ruler and a compass.

Solution

We are given two circles in a plane such that no one is located inside the other (Figure 1a).

We need to construct the common exterior tangent line to the circles using a ruler and a compass.

First, let us analyze the problem and make a sketch (Figures 1 a and 1b). Let AB be the common tangent line to the circles we are searching for.

Let us connect the tangent point A of the first circle with its center P and the tangent point B of the second circle with its center Q (Figure 1 a and 1 b).

Then the radii PA and QB are both perpendicular to the tangent line AB (lesson A tangent line to a circle is perpendicular to the radius drawn to the tangent point under the topic Circles and their properties ).
Hence, theradii PA and QB are parallel.
Figure 1a. To the Problem 1

Figure 1b. To the solution of the Problem 1

Figure 1c. To the construction step 3

Next, let us draw the straight line segment CQ parallel to AB through the point Q till the intersection with the radius PA at the point C (Figure 1b). Then the straight line CQ is parallel to AB.

Hence, the quadrilateral CABQ is a parallelogram (moreover, it is a rectangle) and has the opposite sides QB and CA congruent. The point C divides the radius PA in two segments of the length r2 (CA) and r1 - r2 (PC).

It is clear from this analysis that the straight line QC is the tangent line to the circle of the radius r1 - r2 with the center at the point P (shown in red in Figure 1b).
It implies that the procedure of constructing the common exterior tangent line to two circles should be as follows:

1. draw the auxiliary circle of the radius r1 - r2 at the center of the larger circle (shown in red in Figure 1b);
2. construct the tangent line to this auxiliary circle from the center of the smaller circle (shown in red in Figure 1b). In this way you will get the tangent point C on the auxiliary circle of the radius r1 - r2;
3. draw the straight line from the point P to the point C and continue it in the same direction till the intersection with the larger circle (shown in blue in Figure 1 b). The intersection point A is the tangent point of the common tangent line and the larger circle. Figure 1 c reminds you how to perform this step.
4. draw the straight line QB parallel to PA till the intersection with the smaller circle (shown in blue in Figure 1b). The intersection point B is the tangent point of the common tangent line and the smaller circle;
5. the required common tangent line is uniquely defined by its two points A and B.
Note that all these operations 1 - 4 can be done using a ruler and a compass. The problem is solved.

Problem 2

Find the length of the common exterior tangent segment to two given circles in a plane, if they have the radii r1 and r2 and the distance between their centers is d.
No one of the two circles is located inside the other.

Solution

Let us use the Figure 1b from the solution to the previous Problem 1 .

This Figure is relevant to the Problem 2. It is copied and reproduced in the Figure below on the right for your convenience.

It is clear from the solution of the Problem 1 above that the common exterior tangent segment |AB| is congruent to the side |CQ| of the quadrilateral (rectangle) CABQ.
From the other side, the segment CQ is the leg of the right-angled triangle DELTAPCQ. This triangle has the hypotenuse's measure d and the other leg's measure
r1  r2 . Therefore, the length of the common exterior tangent segment |AB| is equal to
|AB| = √(
d− (r r2)2)
Note that the solvability condition for this problem is d > r1 r2.
It coincides with the condition that no one of the two circles lies inside the other.

Example 1

Find the length of the common exterior tangent segment to two given circles in a plane, if their radii are 6 cm and 3 cm and the distance between their centers is 5 cm.

Solution

Use the formula (1 ) derived in the solution of the Problem 2.
According to this formula, the length of the common exterior tangent segment to the two given circles is equal to
√[52 − (6 − 3)2] = √(52 − 32) = √(25 − 9)
= 4 cm

The length of the common exterior tangent segment to the two given circles is 4 cm

### Contact of Circles

• Two circle are said to touch each other at a point if they have a common tangent at that point.
• Point T is shown by the red dot.

Note;

• The centers of the two circles and their point of contact lie on a straight line
• When two circles touch each other internally, the distance between the centers is equal to the difference of the radii i.e. PQ= TP − TQ
• When two circles touch each other externally, the distance between the centers is equal to the sum of the radii i.e. OR =TO +TR

### Transverse (Interior) Common Tangents

In the figure below P and Q are centres of two circles with radiir, and r, respectively. Given that r> r2, construct the transverse common tangents to both circles.

Procedure

1. With centre P, construct a circle whose radius PR is equal to r+ r2
2. Join P to Q and bisect PQ to get point C.
3. With centre C and radius PC, draw arcs to cut the circle whose radius is r1 + r2, at R and S. Join to R and Q to S.
4. Draw the lines PR and PS to cut the circle whose radius is r1 at M and N respectively.
5. Draw line QX parallel to PS and line QY parallel to PR.
6. Draw lines MY and NX. These are the required transverse (interior) common tangents.

Note:
PR = r1 + r2, (construction).
PM =r1 (given)
∴ RM = PR − PM (r1 + r2) − r1
∴ RM QY But RM is parallel to QY (construction)
∴ MRQY is a parallelogram (opposite sides equal and parallel).
QR is tangent to circle radius PR (construction).
∠PRO = 90° (radius ⟂ tangent).
But ∠YQR = 90° (interior ∠s of a parallelogram).
∴ MRQY is a rectangle.
∴ ∠PMY = ∠QYM = 90°.

### Alternate Segment Theorem

The angle which the chord makes with the tangent is equal to the angle subtended by the same chord in the alternate segment of the circle.

Angle a = Angle b

Note;
The blue line represents the angle which the chord CD makes with the tangent PQ which is equal to the angle b which is subtended by the chord in the alternate segment of the circle.

Illustrations

• Angle s = Angle t
• Angle a = Ange b we use the alternate segment theorm

### Tangent-secant Segment Length Theorem

If a tangent segment and secant segment are drawn to a circle from an external point, then the square of the length of the tangent equals the product of the length of the secant with the length of its external segment.

(TV)
2 = TW.TX

Example

In the figure above ,TW=1 0 cm and XW = 4 cm. find TV

Solution

(TV)2 = TW.TX
(TV)
2 = 10 ×6 ( tx = tw − xw)
= 16
TV = 4 cm

## Circles and Triangles

### Inscribed Circle

• Construct any triangle ABC.
• Construct the bisectors of the three angles
• The bisectors will meet at point I
• Construct a perpendicular from O to meet one of the sides at M
• With the centre I and radius IM draw a circle
• The circle will touch the three sides of the triangle ABC
• Such a circle is called an inscribed circle or in circle.
• The centre of an inscribed circle is called the incentre

### Circumscribed Circle

• Construct any triangle ABC.
• Construct perpendicular bisectors of AB , BC, and AC to meet at point O.
• With O as the centre and using OB as radius, draw a circle
• The circle will pass through the vertices A , B and C as shown in the figure below

### Escribed Circle

• Construct any triangle ABC.
• Extend line BA and BC
• Construct the perpendicular bisectors of the two external angles produced
• Let the perpendicular bisectors meet at O
• With O as the centre draw the circle which will touch all the external sides of the triangle

Note;

• Centre O is called the ex-centre
• AO and CO are called external bisectors.

## Past KCSE Questions on the Topic.

1. The figure below represents a circle a diameter 28 cm with a sector subtending an angle of 750 at the centre.

Find the area of the shaded segment to 4 significant figures
1. ∠PST
2. The figure below represents a rectangle PQRS inscribed in a circle centre 0 and radius 17 cm. PQ = 16 cm.

Calculate
1. The length PS of the rectangle
2. The angle POS
3. The area of the shaded region
3. In the figure below, BT is a tangent to the circle at B. AXCT and BXD are straight lines. AX = 6 cm, CT = 8 cm, BX = 4.8 cm and XD = 5 cm.

Find the length of
1. XC
2. BT
4. The figure below shows two circles each of radius 7 cm, with centers at X and Y. The circles touch each other at point Q.

Given that ∠AXD = ∠BYC = 1200 and lines AB, XQY and DC are parallel, calculate the area of:
1. Minor sector XAQD (Take π 22/7)
2. The trapezium XABY
5. The figure below shows a circle, centre, O of radius 7 cm. TP and TQ are tangents to the circle at points P and Q respectively. OT =25 cm.

Calculate the length of the chord PQ
6. The figure below shows a circle centre O and a point Q which is outside the circle

Using a ruler and a pair of compasses, only locate a point on the circle such that angle OPQ = 90o
7. In the figure below, PQR is an equilateral triangle of side 6 cm. Arcs QR, PR and PQ arcs of circles with centers at P, Q and R respectively.

Calculate the area of the shaded region to 4 significant figures
8. In the figure below AB is a diameter of the circle. Chord PQ intersects AB at N. A tangent to the circle at B meets PQ produced at R.

Given that PN = 14 cm, NB = 4 cm and BR = 7.5 cm, calculate the length of:
1. NR
2. AN

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