Basic Properties of Circles (1)

7

7.1 Chords of a Circle

7.2 Angles of a Circle

Chapter Summary

Case Study

Basic Properties of Circles (1)

7.3 Relationship among the Chords, Arcs and Angles of a Circle

7.4 Basic Properties of a Cyclic Quadrilateral

P. 2

In order to find the centre of the circular plate:

Step 1: Draw an arbitrary triangle inscribed in the circular plate.

Case Study

You need to find the centre of the circular plate first.

I found a fragment of a circular plate. How can I know its original size?

Step 2: Find the circumcentre of the triangle, i.e., the centre of the circular plate, by drawing 3 perpendicular bisectors.

P. 3

A. Basic Terms of a CircleA. Basic Terms of a Circle7.1 Chords of a Circle7.1 Chords of a Circle

Circle: closed curve in a plane where every point on the curve is equidistant from a fixed point.

Centre: fixed point

Circumference: curve or the length of the curve

Chord: line segment with two end points on the circumference

Radius: line segment joining the centre to any point on the circumference

Diameter: chord passing through the centre

Remarks:1. The length of a radius is half that of a diameter.2. A diameter is the longest chord in a circle.

P. 4

A. Basic Terms of a CircleA. Basic Terms of a Circle7.1 Chords of a Circle7.1 Chords of a Circle

Arc: portion of the circumference

Angle at the centre: angle subtended by an arc or a chord at the centre

minor arc (e.g. AYB)shorter than half of the circumference

(

major arc (e.g. AXB)longer than half of the circumference

(

P. 5

B. Chords a CircleB. Chords a Circle7.1 Chords of a Circle7.1 Chords of a Circle

If we draw a chord AB on a circle and fold the paper as shown below:

Then the crease passes through the centre of the circle;

bisects the chord AB. is perpendicular to the chord AB;

A, B coincide

P. 6


Properties about a perpendicular line from the centre to a chord:1. Perpendicular Line from Centre to a Chord

Theorem 7.1If a perpendicular line is drawn from a centre ofa circle to a chord, then it bisects the chord. In other words, if OP AB,

then AP BP. (Reference: line from centre chordbisects chord)

This theorem can be proved by considering AOP and BOP.

P. 7


This theorem can be proved by considering OAP and OBP.

Theorem 7.2If a line is joined from the centre of a circleto the mid-point of a chord, then it isperpendicular to the chord. In other words, if AP BP,

then OP AB. (Reference: line from centre to mid-pt. of chord chord)

The converse of Theorem 7.1 is also true.

P. 8


From Theorem 7.1 and Theorem 7.2, we obtain an important property of chords:

The perpendicular bisector of any chord ofa circle passes through the centre.

P. 9

Example 7.1T

Solution:

In the figure, O is the centre of the circle. AP PB 5 cm and OP 12 cm. Find PQ.


∵ AP PB ∴ OP AB (line from centre to mid-pt. of chord chord)

In OAP,OA2 OP2 AP2 (Pyth. Theorem)

OA cm22 512 13 cm OQ OA (radii)

13 cm∴ PQ (13 – 12) cm

1 cm

P. 10

Example 7.2T

Solution:

In the figure, O is the centre of the circle. AOB is a straight line and OM BC. Show that ABC OBM.


∵ OM BC ∴ BM MC (line from centre chord bisects chord)∴ BC : BM 2 : 1

∵ OB OA (radii)∴ AB : OB 2 : 1OBM ABC (common )

∴ ABC OBM (ratio of 2 sides, inc. )

P. 11

Example 7.3T

Solution:

In the figure, O is the centre and AB is a diameter of the circle. AB CD, PB 4 cm and CD 16 cm. (a) Find the length of PC. (b) Find the radius of the circle.


In OCP,OC2 OP2 PC2 (Pyth. Theorem)

(a) ∵ OB CD ∴ PC PD (line from centre chord bisects chord) 8 cm

(b) Let r cm be the radius of the circle.Then OC r cm and OP (r – 4)

cm.

r2 (r – 4)2 82 8r 80

r 10 ∴ The radius of the circle is 10 cm.

P. 12


Properties about a perpendicular line from the centre to a chord:2. Distance between Chords and Centre

This theorem can be proved by considering OAP and OCQ.

Theorem 7.3If the lengths of two chords are equal, thenthey are equidistant from the centre. In other words, if AB CD,

then OP OQ. (Reference: equal chords, equidistantfrom centre)

P. 13


This theorem can be proved by considering OAP and OCQ.

The converse of Theorem 7.3 is also true.

Theorem 7.4If two chords are equidistant from the centreof a circle, then their lengths are equal. In other words, if OP OQ,

then AB CD. (Reference: chords equidistant from centre are equal)

P. 14

Example 7.4T

Solution:

In the figure, O is the centre of the circle. AB CD, AB CD, OM AB and ON CD. If OP 6 cm, find ON. (Give the answer in surd form.)


∵ AB CD ∴ OM ON (equal chords, equidistant from centre)

ON cm262

∵ All of the interior angles of the quadrilateral OMPN are right angles and OM ON.

∴ OMPN is a square.In ONP,

OP2 ON2 NP2 (Pyth. Theorem) 2ON2

cm

23

P. 15

A. The Angle at the CircumferenceA. The Angle at the Circumference7.2 Angles of a Circle7.2 Angles of a Circle

Angle at the circumference:angle subtended by an arc (or a chord) at the circumference

Angle at the centre:angle subtended by an arc (or a chord) at the centre

Relationship between these angles:

Theorem 7.5

In each of the above figures, the angle at the centre subtended by an arc is twice the angle at the circumference subtended by the same arc. This means that 2. (Reference: at the centre twice at ⊙ce)

P. 16


This theorem can be proved by constructing a diameter PQ.

Since OA OP (radii), AOP is isosceles.∴ OAP OPA a.Hence the exterior angle of AOQ 2a.

In the left semicircle:

Similarly, in the right semicircle, BOQ 2b.

∵ 2a 2b and a b ∴ 2

P. 17

Example 7.5T

Solution:

In the figure, AB and CD are two parallel chords of the circle with centre O. BOD 70 and MDO 10. Find ODC.

∵ BOD 2 BAD ( at the centre twice at ⊙ce) ∴ BAD 35


ODC 10 BAD (alt. s, AB // CD) ODC 10 35 ODC 25

P. 18

B. The Angle in a SemicircleB. The Angle in a Semicircle7.2 Angles of a Circle7.2 Angles of a Circle

In the figure, if AB is a diameter of the circle with centre O, then the arc APB is a semicircle and APB is called the angle in a semicircle.

Since the angle at the centre AOB 180, the angle at the circumference APB 90.

( at the centre twice at ⊙ce)

Theorem 7.6The angle in a semicircle is 90. That is, if AB is a diameter,

then APB 90. (Reference: in semicircle)

Conversely, if APB 90,then AB is a diameter.

(Reference: converse of in semicircle)

P. 19

Example 7.6T

Solution:

In the figure, AP is a diameter of the circle with centre O and AC BC. If PCB 50, find(a) PBC and(b) APC.

(a) Since AP is a diameter, ACP 90. ( in semicircle)

7.2 Angles of a Circle7.2 Angles of a Circle

PBC 20

B. The Angle in a SemicircleB. The Angle in a Semicircle

∵ AC BC∴ PAC PBC (base s, isos. )

In ACB,

PAC PBC ACB 180 ( sum of ) 2PBC (90 50) 180

(b) APC PBC PCB (ext. of ) 70

P. 20

C. Angle in the Same SegmentC. Angle in the Same Segment7.2 Angles of a Circle7.2 Angles of a Circle

Segment:region enclosed by a chord and the corresponding arc subtended by the chord Major segment APB

area greater than half of the circle Minor segment AQB

area less than half of the circle

Angles in the same segment:angles subtended on the same side of a chord at the circumference

Notes:We can construct infinity many angles in the same segment.

P. 21

C. Angle in the Same SegmentC. Angle in the Same Segment7.2 Angles of a Circle7.2 Angles of a Circle

Theorem 7.7The angles in the same segment of a circle are equal, that is, if AB is a chord,

then APB AQB. (Reference: s in the same segment)

The angles in the same segment of a circle are equal.

This theorem can be proved by considering the angle at the centre.

P. 22

Example 7.7T

Solution:

In the figure, AC and BD are two chords that intersect at P. (a) Show that ABP DCP. (b) If AP 8, BP 12 and PC 6, find PD.

(a) In ABP and DCP,

7.2 Angles of a Circle7.2 Angles of a Circle

A D (s in the same segment)

(b) ∵ ABP DCP

PD 4

C. Angle in the Same SegmentC. Angle in the Same Segment

B C (s in the same segment)APB DPC (vert. opp. s)

∴ ABP DCP (AAA)

PCPB

PDPA

∴ (corr. sides, s)

6128

PD

P. 23

7.3 Relationship among the Chords,7.3 Relationship among the Chords, Arcs and Angles of a Circle Arcs and Angles of a Circle1. Equal Chords and Equal Angles at the Centre

Theorem 7.8In a circle, if the angles at the centre are equal,then they stand on equal chords, that is,

if x y,then AB CD.

(Reference: equal s, equal chords)

Conversely, equal chords in a circle subtendequal angles at the centre, that is,

if AB CD,then x y.

(Reference: equal chords, equal s)

This theorem can be proved using congruent triangles.

P. 24

7.3 Relationship among the Chords,7.3 Relationship among the Chords, Arcs and Angles of a Circle Arcs and Angles of a Circle2. Equal Angles at the Centre and Equal Arcs

Theorem 7.9In a circle, if the angles at the centre are equal,then they stand on equal arcs, that is,

if p q,

then AB CD. (Reference: equal s, equal arcs)

Conversely, equal arcs in a circle subtendequal angles at the centre, that is,

if AB CD,then p q.

(Reference: equal arcs, equal s)

( (

( (

P. 25

7.3 Relationship among the Chords,7.3 Relationship among the Chords, Arcs and Angles of a Circle Arcs and Angles of a Circle3. Equal Chords and Equal Arcs

Theorem 7.10In a circle, equal chords cut arcs with equallengths, that is,

if AB CD,

then AB CD. (Reference: equal chords, equal arcs)

Conversely, equal arcs in a circle subtendequal chords, that is,

if AB CD,then AB CD.

(Reference: equal arcs, equal chords)

( (

( (

P. 26

7.3 Relationship among the Chords,7.3 Relationship among the Chords, Arcs and Angles of a Circle Arcs and Angles of a CircleThe above theorems are summarized in the following diagram:

Equal Chords

Equal Arcs Equal AnglesTheorem 7.10

Theorem 7.8Theore

m 7.9

Example: In the figure, the chords AB, BC and CA are of the same length. ∴ Each of the angles at the centre are equal,

i.e., AOB BOC COA 120. ∴ Each of the arcs are equal,

i.e., AB BC CA 9 cm.

(((

P. 27

Example 7.8T

Solution:(a) ∵ AB BC CD DE EF FA ∴ AOB BOC COD DOE EOF FOA

(equal chords, equal s)

(b) ∵ AB BC CD DE EF FA

7.3 Relationship among the Chords,7.3 Relationship among the Chords, Arcs and Angles of a Circle Arcs and Angles of a Circle

In the figure, O is the centre of the circle with circumference 30 cm. A regular hexagon ABCDEF is inscribed in the circle.(a) Find AOB. (b) Find the length of AB.

(

∴ AOB 360 6

∴ AB BC CD DE EF FA (equal chords, equal arcs)

( ( ( ( ( (

∴ AB (30 6) cm

(

5 cm

60

P. 28

7.3 Relationship among the Chords,7.3 Relationship among the Chords, Arcs and Angles of a Circle Arcs and Angles of a Circle4. Arcs Proportional to Angles at the Centre

Theorem 7.11In a circle, arcs are proportional to the anglesat the centre, that is,

AB : PQ : .(Reference: arcs prop. to s at centre)

( (

Notes:1. In a circle, chords are not proportional to the angles subtend at the centre.2. In a circle, chords are not proportional to the arcs.

P. 29

Example 7.9T

Solution:


In the figure, O is the centre of the circle. APB 15 cm, PB 6 cm and POB 80. Find AOP.

( (

AP 9 cm

(

AOP : POB AP : PB (arcs prop. to s at centre)

( (

AOP : 80 9 cm : 6 cm AOP 120

P. 30

Example 7.10T

Solution:


AOB : BOC AB : BC (arcs prop. to s at centre)

( (

40 : BOC 2 : 3 BOC 60

In the figure, O is the centre of the circle. 3AB 2BC and

AOB 40. Find ABC : AEDC.

(

(

( ∵ 3AB 2BC

( (

∴ AB : BC 2 : 3

( (

∴ AOC 100 and Reflex AOC 260

5 : 13∴ ABC : AEDC 100 : 260 (arcs prop. to s at centre)

(

P. 31

7.3 Relationship among the Chords,7.3 Relationship among the Chords, Arcs and Angles of a Circle Arcs and Angles of a Circle5. Arcs Proportional to Angles at the Circumference

Notes:In a circle, chords are not proportional to the angles subtend at the circumference.

Theorem 7.12In a circle, arcs are proportional to the anglessubtended at the circumference, that is,

AB : PQ a : b.(Reference: arcs prop. to s at ⊙ce)

( (

This theorem can be proved by constructing the corresponding angles at the centre for each arcs.

P. 32

Example 7.11T

Solution:


In AOE, OEC 80 20 (ext. of )

In the figure, O is the centre of the circle. AOB 80, OAC 20 and AB 12 cm.(a) Find CBD. (b) Find the length of CD.

(

(

(a) ∵ AOB 2 ACB ( at the centre twice at ⊙ce) ∴ ACB

40

100 In BCE,OEC ACB CBD (ext. of ) 100 40 CBDCBD 60

12 cm : CD 40 : 60(b) AB : CD ACB : CBD (arcs prop. to s at ⊙ce)

( (

∴ CD 18 cm

P. 33

7.4 Basic Properties of a Cyclic7.4 Basic Properties of a Cyclic Quadrilateral QuadrilateralCyclic quadrilateral:quadrilateral with all vertices lying on a circle

A. Opposite Angles of a Cyclic QuadrilateralA. Opposite Angles of a Cyclic Quadrilateral

Two pairs of opposite angles: BAD and DCB ABC and CDA

Theorem 7.13The opposite angles in a cyclic quadrilateralare supplementary.Symbolically, BAD DCB 180 and

ABC CDA 180.(Reference: opp. s, cyclic quad.)

This theorem can be proved by constructing the corresponding angles at the centre.

P. 34

Example 7.12T

Solution:

In the figure, ABCD is a cyclic quadrilateral. AD is a diameter of the circle and DAC 35. Find ABC.

7.4 Basic Properties of a Cyclic7.4 Basic Properties of a Cyclic Quadrilateral QuadrilateralA. Opposite Angles of a Cyclic QuadrilateralA. Opposite Angles of a Cyclic Quadrilateral

∵ AD is a diameter.∴ ACD 90 ( in semicircle)

In ACD,35 ACD ADC 180 ( sum of )

∴ ABC ADC 180 (opp. s, cyclic quad.)

35 90 ADC 180 ADC 55

ABC 125

P. 35

7.4 Basic Properties of a Cyclic7.4 Basic Properties of a Cyclic Quadrilateral QuadrilateralFrom Theorem 7.13, we obtain the following relationship between the exterior angle and the interior opposite angle of a cyclic quadrilateral:

B. Exterior Angles of a Cyclic QuadrilateralB. Exterior Angles of a Cyclic Quadrilateral

Theorem 7.14The exterior angle of a cyclic quadrilateralis equal to the interior opposite angle,that is, .(Reference: ext. , cyclic quad.)

P. 36

Example 7.13T

Solution:

In the figure, two circles meet at C and D. ADE and BCF are straight lines. If BAD 105, find DEF.

7.4 Basic Properties of a Cyclic7.4 Basic Properties of a Cyclic Quadrilateral Quadrilateral

FCD BAD (ext. , cyclic quad.)

B. Exterior Angles of a Cyclic QuadrilateralB. Exterior Angles of a Cyclic Quadrilateral

105 ∴ DEF FCD 180 (opp. s, cyclic quad.)

DEF 75

P. 37

7.4 Basic Properties of a Cyclic7.4 Basic Properties of a Cyclic Quadrilateral QuadrilateralPoints are said to be concyclic if they lie on the same circle.

C. Tests for Concyclic PointsC. Tests for Concyclic Points

To test whether a given set of 4 points are concyclic (or a given quadrilateral is cyclic):

Theorem 7.15 (Converse of Theorem 7.7)In the figure, if p q,

then A, B, C and D are concyclic.(Reference: converse of s in the same segment)

P. 38

7.4 Basic Properties of a Cyclic7.4 Basic Properties of a Cyclic Quadrilateral QuadrilateralC. Tests for Concyclic PointsC. Tests for Concyclic Points

Theorem 7.16 (Converse of Theorem 7.13)In the figure, if a c 180 (or b d 180),

then A, B, C and D are concyclic.(Reference: opp. s supp.)

Theorem 7.17 (Converse of Theorem 7.14)In the figure, if p q,

then A, B, C and D are concyclic.(Reference: ext. int. opp. )

P. 39

Example 7.14T

Solution:

In the figure, APB and RDQC are straight lines. If AD // PQ, show that P, Q, C and B are concyclic.

7.4 Basic Properties of a Cyclic7.4 Basic Properties of a Cyclic Quadrilateral Quadrilateral

ADR ABC (ext. , cyclic quad.)

∴ ABC PQR

C. Tests for Concyclic PointsC. Tests for Concyclic Points

ADR PQR (corr. s, AD // PQ)

∴ P, Q, C and B are concyclic. (ext. int. opp. )

P. 40

Example 7.15T

Solution:

Consider the cyclic quadrilateral PQCD.(a) Find y.(b) Write down another four concyclic points.

7.4 Basic Properties of a Cyclic7.4 Basic Properties of a Cyclic Quadrilateral QuadrilateralC. Tests for Concyclic PointsC. Tests for Concyclic Points

(a) y 110 180 (opp. s, cyclic quad.)

(b) ∵ ABQ QPD 70 ∴ A, B, Q and P are concyclic. (ext. int. opp. )

y 70

P. 41

7.1 Chords of a Circle

Chapter Summary

1. If a perpendicular line is drawn from the centre of the circle to a chord, then it bisects the chord, and vice versa.

2. If the lengths of two chords are equal, then they are equidistant from the centre of the circle, and vice versa.

P. 42

7.2 Angles of a Circle

Chapter Summary

1. The angle at the centre is twice the angle at the circumference subtended by the same arc, that is, x 2y.

2. If AB is a diameter, thenAPB 90.

Conversely, if the angle at the circumference APB 90, then AB is a diameter.

3. The angles in the same segment are equal, that is, x y.

P. 43

Chapter Summary7.3 Relationship among the Chords, Arcs and Angles of a Circle

1. Equal angles at the centre stand on equal chords.2. Equal angles at the centre stand on equal arcs.3. Equal arcs subtend equal chords.

4. Arcs are proportional to the angles at the centre.

AB : PQ x : y

((

5. The arcs are proportional to the angles subtended at the circumference, that is,

AB : BC x : y.

((

P. 44

7.4 Basic Properties of a Cyclic Quadrilateral

Chapter Summary

If ABCD is a cyclic quadrilateral, then(a) a b 180 and(b) a c.

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Basic Properties of Circles (1)