NOTES ON PLANE GEOMETRY II (CIRCLE THEOREM)
Introduction
A circle is a set of points in a plane, which are at equal distance from a fixedpoint O. The point O is called the Centre of the circle and the circular path of the
circle is also called the circumference. Any diameter is a line of symmetry of the
circle. The diagram below shows the parts of a circle.
Segment
Diameter
D
E
Chord
Sector
O
C
A
Centre
Arc length
B
1. Chord: the chord is a line segment that has both end points on the
circumference of the circle.
2. Radius: a line segment drawn from the Centre to any part of the
circumference of the circle. It is the same as half of the diameter of the
circle.
3. Diameter: any line segment (chord) which passes through the Centre of the
circle and has both ends on the circumference.
4. Arc length: the distance 𝐴𝐵 is called the minor arc and the distance
𝐴𝐸𝐷𝐶𝐵 is called the major arc. The arc length represent part of the
circumference of the circle.
5. Segment: a circular region bounded by an arc and the chord of the circle or
an area cut off by the chord and the arc of the circle. The shaded portion is
called the minor segment and the remaining portion is called the major
segment.
SIR LISBON
1
6. Circumference: it is the total distance around the circle. Thus the perimeter
of the circle is called the circumference.
Theorems of circles
Theorem 1
The angles subtended at the circumference of a circle by the same arc or chord are
equal.
𝑺
𝑻
𝑹
𝑦
𝑥
𝑧
𝑂
𝑷
(a)
𝑸
From the diagram (a) above 𝑃𝑄𝑅𝑆𝑇 are points on the circumference of the circle
with centre O.
< PTQ = < PSQ = < PRQ
<𝑥 = < 𝑦 = < 𝑧
𝑹
𝑧
𝑺
ℎ
𝑦
𝑔𝑓
(b)
𝒙
𝑤
v
𝑸
𝑷
SIR LISBON
2
From the diagram (b) above,
a.
b.
c.
d.
< 𝑃𝑆Q = < PRQ ∴ y = x
< QPR =< Q SR ∴ f = h
< 𝑆𝑅𝑃 =< 𝑆Q P ∴ z = v
< 𝑅Q S =< RPS ∴ w = g
Examples:
𝑺
𝑻
1.
𝑹
65°
𝑷
𝑸
P, Q, R, S are points on a circle and< 𝑃𝑆𝑄 = 65°. Find < 𝑃𝑅𝑄 and < 𝑃𝑇𝑄
SOLUTION
From the circle, PQ is the chord
< 𝑃𝑅𝑄 =< 𝑃𝑆𝑄 =< 𝑃𝑇𝑄
< 𝑃𝑅𝑄 = 65°
< 𝑃𝑇𝑄 = 65°
𝑪
2.
𝑦
𝑫
38°
50°
𝑥
𝑛
42°
𝑩
𝑨
In the diagram above, 𝐴, 𝐵, 𝐶, 𝐷 are points on the circle, < 𝐴𝐵𝐷 = 42°, < 𝐴𝐶𝐵 =
50° 𝑎𝑛𝑑 < 𝐵𝐷𝐶 = 38°. Find the values of 𝑥, 𝑦, 𝑎𝑛𝑑 𝑛
SIR LISBON
3
SOLUTION
< 𝐴𝐷𝐵 𝑎𝑛𝑑 < 𝐴𝐶𝐵 Are produced at the circumference by the chord AB
< 𝐴𝐷𝐵 =< 𝐴𝐶𝐵
𝑥 = 50°
< 𝐷𝐶𝐴 = 𝐷𝐵𝐴
𝑦 = 42°
From ∆𝐷𝐵𝐶,
38° + 𝑦 + 50° + 𝑛 = 180°
130° + 𝑛 = 180°
𝑛 = 180° − 130°
𝑛 = 50°
Theorem 2
The angle which a chord or an arc of a circle subtends at the Centre of the circle is
twice the angle subtended by the same chord or arc at any point on the
circumference.
R
R
𝑦
𝑦
O
O
𝑥
𝑥
P
P
Q
Q
O
𝑥
𝑦
R
P
Q
SIR LISBON
4
In all the three circles (a), (b), and (c), the minor arc PQ subtends < 𝑃𝑂𝑄 at the
centre and < 𝑃𝑅𝑄 at the circumference of the circles.
Note: draw the circles and measure the angles 𝑥 𝑎𝑛𝑑 𝑦 as shown.
1
< 𝑃𝑂𝑄 = 2 < 𝑃𝑅𝑄 Or < 𝑃𝑅𝑄 = < 𝑃𝑂𝑄
2
1
< 𝑥 = 2 < 𝑦 Or < 𝑦 = < 𝑥
2
Examples:
1. P, Q, R, are points on the circumference of the circle with Centre O.
if< 𝑃𝑅𝑄 = 62°, find the value of𝑦.
R
62°
O
𝑦
P
Q
SOLUTION
The angle which the arc PQ subtends at the Centre of the circle is twice that if
subtends at the circumference.
➢ < 𝑃𝑂𝑄 = 2 < 𝑃𝑅𝑄
∴ 𝑦 = 2(62°) = 124°
SIR LISBON
5
2. In the diagram below, 𝑃, 𝑄, 𝑅, 𝑆 are points on a circle with centre O.
If < 𝑃𝑅𝑄 = 68°, find the value of 𝑦.
TRY:
R
S
68°
O
P
Q
Theorem 3
The angle subtended by a diameter at the circumference of the circle is 90°
S
T
𝑦
𝑥
P
O
R
𝑧
Q
The line PR is a diameter of the circle with Centre O. measure all the angles
< 𝑃𝑄𝑅, < 𝑃𝑇𝑅 𝑎𝑛𝑑 < 𝑃𝑆𝑅.
< 𝑃𝑄𝑅 =< 𝑃𝑇𝑅 =< 𝑃𝑆𝑅 = 90°
∴ < 𝑥 =< 𝑦 =< 𝑧 = 90°
SIR LISBON
6
Example
R
32° 𝑛
50°
37°
O
Q
S
In the diagram, PQ is the diameter of the circle with Centre O. calculate < 𝑃𝑄𝑆
and the value of n.
SOLUTION
(a) Since PQ is the diameter of the circle, < 𝑃𝑆𝑄 = 90°
37° + 90°+< 𝑃𝑄𝑆 = 180°
127°+< 𝑃𝑄𝑆 = 180°
< 𝑃𝑄𝑆 = 180° − 127°
< 𝑃𝑄𝑆 = 53°
(b) Since PQ is the diameter of the circle,
< 𝑃𝑅𝑄 = 90°
𝑛 + 32° = 90°
𝑛 = 90° − 32°
𝑛 = 58°
TYR
Find < 𝐵𝐴𝐶 if AB is the diameter of the circle and < 𝐴𝐵𝐶 = 42°
SIR LISBON
7
A
O
42°
B
C
Theorem 4
The opposite angle of a cyclic quadrilateral are supplementary (ie. 180°). A cyclic
quadrilateral is a quadrilateral whose vertices lie on a circle.
Consider the cyclic quadrilateral PQRS in the diagram below
S
𝑥
P
𝑛
𝑚
R
𝑦
𝑧
Q
If one side of a cyclic quadrilateral is produced, the exterior angle formed is equal
to the interior opposite angle.
𝑥 + 𝑦 = 180°
𝑚 + 𝑛 = 180°
𝑦 + 𝑧 = 180°
➢ z=x
SIR LISBON
8
Examples
T
S
𝑧
𝑥
98°
𝑦
P
100°
Q
R
Find the values of 𝑥, 𝑦 𝑎𝑛𝑑 𝑧
SOLUTION
The opposite angles of a cyclic quadrilateral are supplementary angles.
I.
< QST+< QPT = 180°
𝑥 + 98° = 180°
𝑥 = 180° − 98° = 82°
II.
< 𝑃𝑄𝑆+< 𝑆𝑄𝑅 = 180°
𝑦 + 110° = 180°
𝑦 = 180° − 110° = 70°
III.
< 𝑃𝑇𝑆+< 𝑃𝑄𝑆 − 180°
𝑧 + 70° = 180°
𝑧 = 180° − 70° = 110°
Theorem 5 the angle a chord makes with the tangent to a circle is equal to the
angle subtended by the same chord at the circumference of the circle.
Q
𝑦
N
𝑥
O
•
R
𝑧
T
𝑓
P
SIR LISBON
S
9
In the diagram above, PNQ and R are points on the circle with Centre O. PR is the
chord and TS is the tangent, which touches the circle at P. The line PQ is a
diameter of the circle.
Notes: since PQ is the diameter of the circle,
➢ < PRQ = 90°
< 𝑃𝑄𝑅+< 𝑅𝑃𝑄+< 𝑃𝑅𝑄 = 180°
< 𝑦+< 𝑧 + 90° = 180°
< 𝑦+< 𝑧 = 180° − 90°
< 𝑦 + 𝑧 = 90°………………… (1)
Since the tangent TS is perpendicular to the diameter or the radius of the circle,
< 𝑅𝑃𝑄 + 𝑆𝑃𝑅 = 90°
< 𝑧+< 𝑓 = 90°…………………………… (2)
Compare the two equations (1) and (2)
< 𝑧+< 𝑓 =< 𝑦+< 𝑧
< 𝑓 =< 𝑦
∴< 𝑆𝑃𝑅 =< 𝑃𝑄𝑅
From diagram (a)
< 𝑥 =< 𝑦
< 𝑓 =< 𝑥
➢ < PNR =< PQR =< SPR
Example
Find < 𝑇𝑄𝑆 𝑖𝑓 < 𝑄𝑆𝑇 = 60° 𝑎𝑛𝑑 < 𝑆𝑄𝑅 = 72°.
SIR LISBON
10
T
60°
S
72°
P
Q
R
SOLUTION
The angle between the chord QS and the tangent PR is equal to the angle produced
by the same chord QS at T.
< QTS =< SQR
➢ < QTS = 72°
From the triangle QST
< 𝑇QS+< QTS + 60° = 180°
< 𝑇QS + 72° + 60° = 180°
< 𝑇QS = 180° − (72° + 60°)
< 𝑇QS = 48°
Theorem 6 (chord properties)
i.
ii.
iii.
iv.
Equal chords subtend equal angles at the Centre of the circle.
The line from the Centre of a circle perpendicular to a chord bisects the
chord.
Equal chords in a circle are equidistant from the Centre of a circle.
Chords on a circle that are equidistant from the Centre are equal.
SIR LISBON
11
A
𝜃
O
•
𝜃
D
B
C
< 𝐴𝑂𝐵 =< 𝐶𝑂𝐷 𝑠𝑖𝑛𝑐𝑒 |𝐴𝐵| = |𝐶𝐷|
Theorem7 (TANGENT of a circle)
A tangent to a circle is a straight line which meets the circle at one point only. The
tangents to a circle from an external point are equal in length. The tangent is
perpendicular to the radius or the diameter at the circumference of the circle. The
line joining the external point to the Centre of the circle bisects the angle between
the tangents and the radii drawn to the point of tangent.
C
A
x
x
q
q
r
O
r
B
The lines AC and AB are two tangents from the point A, which touches the
circumference of the circle at C and B respectively.
Note: (i) |AC| = |AB|
(ii) ∆𝐴𝐶𝑂 and ∆𝐴𝐵𝑂 are right – angled triangle. ∴< 𝐴𝐶𝑂 =< 𝐴𝐵𝑂 = 90°
(iii) |OC| = |OB| (Radii)
(iv) < 𝐶𝐴𝑂 =< 𝐵𝐴𝑂 and < 𝐴𝑂𝐵 =< 𝐴𝑂𝐶
SIR LISBON
12
Theorem 8
Equal chords or arcs subtend the same angles at the circumference of a circle.
D
C
a
b
B
E
A
F
From the diagram above, AB and EF are equal chords
➢ < ACB =< EDF
< 𝑎 =< 𝑏
SIR LISBON
13
EXERCISE
1.
̅̅̅̅̅ touches the circle MNOP at P and 𝑁𝑃
̅̅̅̅̅ is a
In the diagram below 𝑃𝑄
diameter. < MPQ = 33° and < PMO =67°. Find:
i.
ii.
< MNO
< MPO
M
N
67°
•
33°
P
O
2.
NOT DRAWN TO SCALE
In the diagram below, PQR is a circle with centre O. if < RQO = 46°, find
< RPQ.
P
O
•
R
46°
Q
NOT DRAWN TO SCALE
SIR LISBON
14
3. In the diagram below, MNPR are points on a circle with Centre O. The
reflex angle at O is 204°, < NMO =52°. Find the value of m.
N
204°
•
52°
O
m
M
P
NOT DRAWN TO SCALE
R
4. In the diagram below, A, B, C, D are points on the circumference of a circle
with Centre O. If AO is parallel to DC and O𝐴̂C = 38°, find
i. < ACB
ii. < DBC
D
A
38°
>
>
C
•O
B
NOT DRAWN TO SCALE
SIR LISBON
15
5. In the diagram below, O is the Centre of the circle QRST. < QRT = 42° and
< PQS =124°. Find
S
i. < RSQ
T
ii. < STR
O•
42°
R
124°
P
Q
NOT DRAWN TO SCALE
6. In the diagram below, PS // QR, < PSR =79°, < SPR = 54° and TQ is a
tangent to the circle at p.
Q
P
T
54°
S
79°
R
NOT DRAWN TO SCALE
SIR LISBON
16
7. In the diagram below, A, B, C and D are points on the circle with Centre O.
< 𝑂𝐴𝐷 = 30° 𝑎𝑛𝑑 < 𝐴𝑂𝐶 = 140°. Find the values of a and b
A
D
30°
a
B
𝑏
O
C
NOT DRAWN TO SCALE
̅̅̅̅ 𝑎𝑛𝑑 𝑄𝑅
̅̅̅̅̅
8. In the diagram below, PQRS is a cyclic quadrilateral. 𝑃𝑆
are produced to meet at T. ̅̅̅̅
𝑆𝑅 𝑎𝑛𝑑 ̅̅̅̅
𝑃𝑄 are produced to meet at U.
if < 𝑃𝑆𝑅 = 84° 𝑎𝑛𝑑 < 𝑆𝑅𝑄 = 126°, find 𝑥 𝑎𝑛𝑑 𝑦.
P
84°
Q
𝑦
126°
S
R
𝑥
U
T
SIR LISBON
17
9.
In the diagram below, PQR is a tangent to the circle at Q, < 𝑌𝑄𝑅 =
54° 𝑎𝑛𝑑 < 𝑋𝑌𝑂 = 41°. Find the size of < 𝑋𝑄𝑌.
X
41°
Y
54°
P
R
Q
10.In the diagram below, POR is a diameter, < 𝑄𝑃𝑅 = 𝑎° 𝑎𝑛𝑑 < 𝑃𝑅𝑆 =
(4𝑎 + 12)°. find the value of a.
P
S
O
R
Q
SIR LISBON
18