### Finding the centre and radius of a circle - Sherman Kashyap and Sheares

Finding the centre
-First of all, we drew two tangents that are not directly opposite each other. Since tangents and radii form 90 degree angles, we can find the diameter of the circle. From there we can either split the diameter into two or find the intersection of the lines perpendicular to the tangent.

-We measure the diameter and split it into half. (In this case the circle we constructed had a radius of 6 cm

### Finding the centre of the circle — Aziel and Charlene

Finding the centre of the circle
We drew two long arcs with the same radius from opposite sides of the circle, making a line that passes through where the arcs intersected. This finds the perpendicular bisector of the undrawn line connecting the two points the arcs originated from.
We turned the paper a rough 90° and repeated the same process. The instance where the two perpendicular bisectors intersect was the centre of the circle.

Proof of how we found the centre of the circle
The perpendicular bisector of any chord in a circle would be its diameter if extended from one point of the circumference to the other. This is because a line connecting the centre of the circle to a chord is perpendicular to the chord it connects the centre to.
What we have done is find the perpendicular bisector of two chords and, since the diameters would intersect in the centre, hence have used this method to find it.

Finding the diameter and radius of the circle
Extending or limiting the line drawn by the arc intercepts to exactly at the circumference and measuring this line, we found that the diameter was approximately 10.25 cm and that the radius was approximately 5.125 cm.

### Finding the centre of a circle - Esther, Dawn & Dylaine

To find the centre of the circle, we drew a chord of the circle, then drew an arc from each of the two end points of the chord of the same radius length. From the two intersecting points formed by the arcs, we were able to construct a diameter of the circle, from the Bisecting Chord Theorem (perpendicular bisector of chord is diameter).

This was repeated to find another diameter, in which the intersection with the first diameter is the centre of the circle.

### Finding the center and radius of a circle (Guo Jin, Luke, Jerome)

1) How to find the center of a circle:

Assume we have a circle

We will use the property: angles in a semicircle = 90º.
This property shows that if a triangle is drawn inside a semicircle, the angle opposite the diameter will be 90º.
With this property in mind, let us draw a 90º angle at a point on the circumference of the circle.

The 2 lines from the 90º angle form 2 sides of a triangle inscribed inside a semicircle. This can be derived by working backwards from the property of angles in a semicircle. Since the triangle is inscribed inside the semicircle, the 3rd side of the triangle is the diameter of the semicircle, and thus the diameter of the circle itself.

We can repeat the process again, and draw another right angle. By extending the lines formed by the right angle, another triangle inscribed in the semicircle is formed. Again, the 3rd side of this triangle is the diameter.

The point where the two diameters intersect is the center of the circle. (All diameters pass through the center, and all diameters intersect only at the center.)

2) How to find the radius of a circle

Method 1
Assuming we have another circle

Draw 2 lines of the same length joined at a 90º angle at the circumference. Join the lines with another line.
These lines form the 2 sides of an inscribed triangle in the semicircle (the semicircle is the circle divided in half). The third side of the triangle would be the diameter of the semicircle which the triangle is inscribed in. The semicircle's diameter is the diameter of the circle itself.

Draw another line down from the point where the lines meet. Since the lines are of equal length, the line drawn here will bisect the diameter, forming 2 radii. The line that is drawn down to meet the diameter is also another radius.

Therefore, the radius can be determined.

Other Methods:
If the center has already been found, using methods such as in the previous part,

Draw a line from the center to the circumference, on one side. This line is the radius.
This method can be used for any other circle which the center is already known.

Otherwise, calculation can be used. If the diameter, circumference or area of the circle is known, their respective formulas can be applied to obtain the radius.

Measurement can also be used.

### Circle Circle Circle

ACTIVITY 3A

Post your responses of the following:

• Why is a circle a 'circle' and why it is NOT a triangle?   Describe its characteristics.
• Give at least 2 examples of daily objects, figures, shapes that could be classified as a circle.

### Find my Centre - Authentic Task

ACTIVITY 3

Objective:
This is a collaborative activity to consolidate the learning of angles in a circle viz a viz APPLIED LEARNING MODE.

ICT tools:

You are a mechanical engineer that is supervising a mould making process. You have just been tasked to replicate a circular object in your production line. The products must be precise (congruent of the highest degree) and must be mass produced efficiently at the shortest possible time with minuscule margin of error.

The first task is to produce a mould by which all the objects will be replicated from.

To facilitate the process, you are required to first determine how you would determine the following
1. centre of the circle
2. radius of the circle / sphere
For the above activity, you are required to use the angles properties discovered in the topic.
Post your solution in blog page assigned to you.
Remember time is a factor so you have to plan your time very well.

### Assessment in Term 4 2013

Dear SSTudents,

1. Performance Task 2 (in lieu of Elementary Mathematics)

As mentioned earlier the deadline of the PT2 is 16 September 2013 @ 2359. To date many students have submitted their products with high quality questions and 'proof's.  Effective use of ICT tools (google, Blog, Geogebra, Keynote, Powerpoint, Pretzi etc) have further enhanced the final product.

2. Paper 3 (in lieu of Additional Mathematics)

The assessment information will be as follows:
Date:    23  September  2013 (Monday)
(Please be punctual and ensure you have a heavier meal in the morning)
Time:   3.00 pm to 4.00 pm
Venue: Auditorium
Note that you are required to sit according to your classes and index numbers. The teachers will supervise you on this.

Logistic:
TI84 Graphic Calculator (or other approved GC model)
(no other calculator will be allowed)
Stationery - pen and ruler

3. Information on EOY

All the best - you can do it because we have faith in you but do you!

### Angles in a Circle - MULTIMODAL LEARNING

ACTIVITY 1

Objective:
This is a self directed activity to consolidate the learning of angles in a circle.

ICT tools:
Math Blog for activity specifications
TI-Nspire CAS
tns file will be given at the beginning of session

ACTIVITY 2

Objective:
This is a collaborative activity to consolidate the learning of angles in a circle viz a viz formative assessment.
Error analysis will follow the assessment.

ICT tools:
TI-Nspire CAS
tns -based poll will be used to assess learning.

ACTIVITY 3

Objective:
This is a collaborative activity to consolidate the learning of angles in a circle viz a viz APPLIED LEARNING MODE.

ICT tools:
TI-Nspire CAS

Given the object, determine how you would determine the following
1. centre of the circle
2. radius of the circle / sphere
For the above activity, you are required to use the angles properties discovered in the topic. Do provide the logical reasoning in your answer.

## The Alternate Segment theorem states

An angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

Recall that a chord is any straight line drawn across a circle, beginning and ending on the curve of the circle.
In the following diagram, the chord CE divides the circle into 2 segments. Angle CEAand angle CDE are angles in alternate segments because they are in opposite segments.

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
In terms of the above diagram, the alternate segment theorem tells us that angleCEA and angle CDE are equal.

Example:
In the following diagram, MN is a tangent to the circle at the point of contact A.Identify the angle that is equal to x

Solution:
We need to find the angle that is in alternate segment to x.
x is the angle between the tangent MN and the chord AB.
We look at the chord AB and find that it subtends angle ACB in the opposite segment.

So, angle ACB is equal to x.

cyclic quadrilateral is a quadrilateral that can be inscribed in a circle. They have a number of interesting properties.

## Applicable Theorems/Formulae

The following theorems and formulae apply to cyclic quadrilaterals:

## Class based activity:

Objectives:
to proof the validity of the following theorems through geometrical manipulations using Geogebra.
Theorem 1: Angle at the centre theorem
Theorem 2: Angle in a semi circle theorem
Theorem 3: Angles in the same segment

Activity 1:
Using the attached links explore the following:
a. the various permutations of Angle at the centre theorem
b. the relationship between Angle at the centre theorem and Angle in a semi circle theorem

Activity 2:
Using the attached links explore the limitations of Angles in the same segment theorem.

### Circle - Chord Properties

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### Essential GDC Skills

Essential GDC Skills

View each video carefully and learn these fundamental GDC skills.

1. Basic Graphing Controls
a. Zoom Options

b. Setting the Window

2. Graphing Basics

a. Graph a line and find the table of values:

b. Finding coordinates of turning points of a graph:

c. Finding intersection between two graphs:
(comes with exercises)

d. Finding roots and y-intercept of a graph:

3. PlySmlt2
a. Using PlySmlt2 for Solving Quadratic Equations

b. Using PlySmlt2 to Solve a System of Equations

c. Using PlySmlt2 to Solve Polynomial Equations