How Important are Circles?

Geometry is considered as a very important field of life because it is also used in our daily life. Circles are one of the main part of geometry. Below are some of the applications of circles in daily life are as follows:

The properties of circles are widely applied in construction of buildings and houses. Secondly, it is used in measurements with the help of rulers in creating maps. Thirdly, circles are very important in making building designs. These properties are also used to understand the circular tires on car. It is also used in designing playgrounds.

The construction of tunnels are closely related to circles. As we can see, tunnels are usually cylindrical in form. This is because tunnels are usually built underground or built through enormous mountains and it must be able to withstand the great amount of weight pressured upon it. Hence, applying the properties of circles is important for us to construct a perfect circle so that the constructed tunnels are more stable and safe to be used.

Tunnel Lining Formwork:

Besides, properties of circles doesn't only mean to construct structures that are circular in form. By using circle, we could acurately construct different kinds of polygons. With the help of circles, designs of products or buildings could be more and more creative and attractive.

The Pentagon

KLCC

Shanghai Tower

Often, properties of circle are also used in calculations and measurements.

Example:

RF Transmitters

It is a method using circles to determine a position of an object instead of x-y coordinates.

Transmitter Configuration

Circles are amazing, aren't they?

Work Examples

Some Work Examples
Qn1)In the figure, CAD and CBE are straight lines. If CA is the diameter of the circle ABC, explain why Angle ADE is a right angle.

Qn2)
ABCD is a cyclic quadrilateral and BC = CD. Show that AC bisects triangle BAD.

Properties of Circles Part 4

Tangent from Exterior point


In the figure on the above, 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.

The alternate segment theorem states that an angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Thus, ÐPTB = ÐPQT.

Alternate Segment Theorem


The alternate segment theorem states that an angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Thus, ÐPTB = ÐPQT.

Proof:

ÐSPT = 90° (rt. Ð in a semicircle)
Ðx + Ðy = 90° ( Ð sum of ∆)
Ðy + Ðz = 90° ( tan rad.)
Ðx = Ðz
ÐPTB = ÐPST
ÐPST = ÐPQT ( Ðs in same segment)
ÐPTB = ÐPQT

Example 1:

In the figure shown, PA and PB are tangents to the circle. ABP = 60° and BAC = 40°. Find the value of x and y.

x° = 60° (Ð in alt. segment)
ÐPAB = 40° (base Ð of isos. ∆PAB)
y° = 180° - 60° - 40° (adj. Ðs on str. l)
= 80°

x is 60 and y is 80.

Properties of Circles Part 3

Cyclic Quadrilaterals

A quadrilateral with its four vertices lying on the circumference of a circle is called a cyclic quadrilateral.



Property 1

In a cyclic quadrilateral, the opposite angles are supplementary,

Proof:

Cyclic Quadrilaterals

Let b = 50

2d = 360 - 100 = 260
d = 260/2 = 30
b + d = 30 + 50 = 180.

Property 2

If one side of a cyclic quadrilateral is produced, the exterior angle so formed is equal to the interior angle.

Proof


b + d = 180°(opp. angles of cyclic quad.)
x + d = 180°(adj. angles on a str. l)
b + d = x + d
b = x
angle ABC = angle CDE

Properties of Circles Part 2

Property 1

An angle at the centre of a circle is twice any angle at the circumference subtended by the same arc.


Proof:

In the figure below, the angles are subtended by the minor arc AB.
Since OA = OD (radii of circle), a = b (base angles of isos. triangle)
But angle AôE if the exterior angle of triangle AOD
AôE = 2a
Similarly, c = d (base angles of isos. triangle)
BôE = 2c
Hence, AôB = 2a + 2c = 2(a + c) = 2 angle ADB

Property 2

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

Proof

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

Property 3

Angles in the same segment of a circle are equal.

Proof


AôB = 2x1 = 2x2 (Ð at centre= 2Ðat ⊙ce)
x1 = x2
ÐAPB = ÐAQB

Done by:Ng Yee Hang(38)

Properties of Circles Part 1

Properties Of circles
Some Definitions:

Center-A point inside the circle. All points on the circle are equidistant (same distance) from the center point.
Radius-The radius is the distance from the center to any point on the circle. It is half the diameter.
Diameter-The distance across the circle. The length of any chord passing through the center. It is twice the radius.
Circumference-The circumference is the distance around the circle.
Chord-A line segment linking any two points on a circle.
Tangent-A line passing a circle and touching it at just one point.
Segment-the region enclosed by a chord and the circumference.
(Note the smaller area under the line”Segment” is the minor segment while the larger area is the Major Segment.)

Property 1:

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.
Proof:

Given a circle, centre O and a chord, AB, with a mid-point D, we are required to show that OĈB = 90°.
Join OA and OB. In triangle OAC and OBC,
OA = OB (radii of circle)
AC = BC (given)
OC is common.

Triangle OCD is congruent to triangle OBC (SSS property)

OĈA = OĈB.

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

Property 2

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

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.


Done by:Ng Yee Hang(38)