Angle in a semicircle


Every angle at the circumference subtended by the diameter of a circle is a right angle triangle.

Property 2 can be abbreviated as rt. Ðin a semicircle.







Proof

AôB = 2AĈB ( at centre = 2 at )
But AôB = 180°
AĈB = 90°


Every angle subtended at the circumference by the diameter of a circle is a right angle (90˚).











POQ is the diameter. ∠PAQ = ∠PBQ = ∠PCQ = 90˚.
Example:











O is the centre of the circle. Find the value of x
Solution:
∠ABC = 90˚ ( angle in a semicircle = 90˚)
63˚ + 90˚ + x = 180˚ ( sum of angles in a triangle )
x = 27˚

3. Equal chords, equidistant from centre. Equal chords of a circle are equidistant from the centre of the circle. In equal circles or in the same circle, equal chords are equidistant from the centre. Chords which are equidistant from the centre are equal.











Proof:
1. In the figure, triangle OAB is rotated through an angle AOA' to triangle OA'B' about O.

Since rotation preserves shape and size, AB = A'B' and OG = OH.













2. In the diagram, AB = CD
Then OM = ON. (equal chords, equidistant from centre)













Example:

In the diagram, O is the centre of the circle, AB = CD and OM = 10cm. Find the length of ON.













Solution:

AB = CD(given)
Therefore, ON = OM (equal chords, equidistant from centre)
= 10 cm

Tangents from External Point
Example 1
In the figure on the right, P is a point outside the circle, with centre O, PA and PB are two tangents drawn from P to touch the circle at A and B respectively. We can find that

i) AP = BP
ii) ÐAPO = ÐBPO
iii) ÐAOP = ÐBOP












ÐOAP = ÐOBP = 90° (tan ⊥ rad.)
△AOP and △BOP are congruent (RHS Property)
AP = BP
ÐAPO = ÐBPO and ÐAOP = ÐBOP

We can conclude that:

a) tangents drawn to a circle from an external point are equal

b) the tangents subtend equal angles at the centre

c) the line joining the external point to the centre of the circle bisects the angle between the tangents.

Example 2

In the figure, AB is a tangent to the circle, with centre O. Given that AB = 8cm, BC = 5cm and OA = x cm, find
(a) the value of x b) ÐAOB
(c) the are bounded by AB, BC and the arc AC.

(a) ÐOAB = 90° (tan ⊥ rad.)
OB = (x + 5)cm















(x + 5)2 = x2 + 82
x2 + 10x + 25 = x2 + 64
10x = 64 - 25 = 39
x = 3.9

(b) tan ÐAOB = 8/3.9
ÐAOB = 64.0° (1 d.p.)

(c) Area AOB = ½(8)(3.9) cm2 = 15.6cm2
Area minor sector AOC = 64.01/360 x p(3.9)2 cm2 = 8.496
= 8.50 cm (3.s.f.)
Area bounded = (15.6 - 8.496)cm2
= 7.10cm2 (3 s.f.)

A circle itself is a curve. Therefore, at each point on the circumference of a circle, we can draw a tangent to touch the circle at that point. For example, TAN is the tangent to the circle at A.

By the symmetry of the diagram,
Angle OAT = Angle OAN
Angle OAT + Angle OAN = 180 degrees (adj. angles on a straight line)
Angle OAN = Angle OAT = 90 degrees

Hence, we have the following property.
A tangent to a circle is perpendicular to the radius of the circle drawn from the point of contact. (Abbreviation: tangent perpendicular to radius)

Example 1:

In the diagram, O is the centre of the of the circle with points B, P and A touching the sides of the circle. TN is the tangent to the circle at A and angle BAT = 35 degrees.
Find angle AOB and angle APB.

















Angle OAT = 90 degrees (tangent perpendicular to radius)
Angle OAB = 90 degrees – 35 degrees
= 55 degrees
OA = OB (radii of the same circle)
Angle OAB = Angle OAB (base angles of isosceles triangle)
= 55 degrees
Angle AOB = 180 degrees – 55 degrees – 55 degrees
= 70 degrees

Angle APB = Angle AOB divided 2 (angle at centre = 2 angles at circumference)
= 70 degrees divided 2
= 35 degrees

The angle subtended by an arc at the centre of a circle is twice the angle subtended by the same arc at the circumference. (Abbreviation: angle at centre = 2 angles at circumference)

Example 1:

In a diagram, O is the centre of the circle with points A, P, B touching the sides of the circle. Angle APB = 47 degrees. Find angle AOB.













Angle AOB = 2 angle APB (Abbreviation: angle at centre = 2 angles at circumference)
= 2 x 47 degrees
= 94 degrees

Example 2:

In a diagram, O is the centre of the circle with points A, P, B touching the sides of the circle. Angle AOB = 110 degrees. Find angle APB.















Angle AOB = 2 angle APB (Abbreviation: angle at centre = 2 angles at circumference)
= 110 degrees divided 2
= 55 degrees

Example 3:

In a diagram, O is the centre of the circle with points A, P, B touching the sides of the circle. Angle AOB = 118 degrees. Find angle APB.

Reflex angle AOB = 360 degrees – 118 degrees (angles at a point)
= 242 degrees
Reflex angle AOB = 2 angles APB
242 degrees = 2 angles APB
Angle APB = 121 degrees

2. Angles in opposite segments. When a quadrilateral is drawn inside a circle and their points touch the sides of the circle, the point at the top and the point at the bottom adds up to be 180°. This is called angles in opposite segments.













In this case, P and Q add up together to become 180°. In full: the sum of the angles in the opposite segments of a circle is 180°.Abbreviation: ∠s in oppo. segments Example:














∠ACB=90° (∠in a semicircle)∠y=180-90-60 =30° (∠s in oppo. segments)

Angles in segments

1. Angles in the same segment. When two triangles are inside a circle and their points touch the sides, angles on the upper layer are the same. This is called angles in the same segment.













In this case, angle A and B is the same because of the property: angle in the same segments.

In full: the angles in the same segment of a circle are equal.
Abbreviation: angles in the same segment.


Examples:
In the diagram, AC and BD intersect at E, ∠ACB=54° and ∠ABD=53°. Find ∠ADB and ∠DAB.












∠ADB=54° (∠s in the same segment)

∠DAB=180-53-54
=73° (sum of ∠ in a triangle)

2. Perpendicular from centre bisects chord
The perpendicular from the centre of a circle to a chord bisects the chord.












In the diagram, ON perpendicular AB
Then AN=BN


Example:
In the diagram, AB and CD are two parallel chords of a circle, centre O. AB=30 cm, CD= 48cm and the perpendicular distance from O to AB is 20cm. Find
a) the radius of the circle,
b) the perpendicular distance between AB and CD














Solutions:
a) Draw the radii OA and OC. Draw the perpendicular ON from O to CD. Since AB//CD, MON is a straight line.

In triangle OAM,

AM=1/2 AB
= 1/2 x 30
= 15cm (perpendicular from centre bisects chord)

OA square = AM square + OM square
OA = Square root 625
= 25cm (pyth thm)

The radius of the circle is 25cm.

a) In triangle OCN,
CN = 1/2CD
= 1/2 x 48
= 24cm

OC square = CN square + ON square
25 square = 24 square + ON square
ON square = 49
ON = 7cm (pyth thm)

Perpendicular distance between AB and CD = MO + ON
= 20 + 7
= 27cm

Chords of circle:

1. Perpendicular bisector of chord
The Perpendicular bisector of chord of a circle passes through the centre of the circle.

A circle is symmetrical about every diameter. Hence any chord AB perpendicular to a diameter is bisected by the diameter.
Also, any chord bisected by a diameter is perpendicular to the diameter.














This property enables us to locate the centre of a circle. The diagram bellows shows a tracing of a broken circular plate. We locate the centre of the plate by marking three points A, B and C on it and drawing the perpendicular bisector of AB and BC. The intersecting point of the perpendicular point of the perpendicular bisectors is the centre.














Proof:

Given a circle, centre O and a chord, SR, with a mid-point Z, we are required to show that OZR = 90°.

ZJoin OS and OR. In triangle OSZ and ORZ,
OS = OR (radii of circle)
SZ = RZ (given)
OZ is common.

Triangle OZS is congruent to triangle OZR (SSS property)

OZR = OZR

Since these are adjacent angles on a straight line, OZR = OZR = 90°

Example:
In the diagram, M is the midpoint of the chord AB and PQ is the perpendicular bisector of AB. AB=8 cm and MQ=2cm. Find the radius of the circle.













Let r be the radius of the circle.

The centre O lies on the line PQ. (Perpendicular bisector of chord)

Join OB as shown in the diagram.

OM square= OM square + MB square

r square =(r-2) square + 4 square
r square = r square – 4r + 4 + 16
4r = 20
r = 5
The radius of the circle is 5 cm.