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Name of School:

ROUTINE INFORMATION Inhlakanipho High School.

Student Surname and Name:

Maphumulo Mthokozisi

Student Number:

207503878

Grade:

11

Subject:

Mathematics

Topic:

Equation of a circle whose centre is not at origin

Content /Concept Area:

-Circles centred off the origin

CAPS page number:

188

Duration of Lesson:

45 minutes

Specific Aims: To develop an understanding of standard equation of a circle. To explore properties of a circle in an in-depth manner. To get learners exposed to, trinomial, polynomial and quadratic function. To expand learners understanding of algebraic function.

Lesson Objectives:

Learners should gain

Knowledge

Skills Learners should be able to do

Value /Attitude Learners should acquire

knowledge and

following :

values and attitudes conducive to :

understanding of the following : Determine the shape of the

Determine

graph using a grid pattern

intercept

the

x-

and

y-

worksheet or transparency

Acknowledge how an equation of a circle is produced in Cartesian plane, where straight a distance formula is used to express an equation of a circle.

Define formula of a circle

Draw the table for x-values and

Acknowledge understanding of


graph

substitute the them to the given

to main types of the equation

equation of the circle , where

of the circle including

x- values corresponds to y-

1) Origin centred circle

values in a Cartesian plane

equation 2) And off origin centred circle equation

Determine, calculate, draw and

Acknowledge students ability

Devise the effective use and

manipulate the equation of the

to explore diversified ways and

application of circle formula

circle using distance formula

resources used to devise circle

and midpoint formula.

equation.

Approaches / Teaching Strategies:

Role plays. Case studies Class discussions. Questioning. Whole class-discussion Group Work Resources:

OHP/ Transparency Calculators Chalkboard Newspaper Worksheet Lesson Phases:

Introduction:

To kick-start this lesson I will ask students to define circle. Students are expected to define a circle with respect to its commonly known properties such as radius, perimeter and areas. To


explore their knowledge possession about a circle I will also bring chalkboard instruments and physical object to demonstrate a circle. This time around we are using a protractor to measure a semi-circle and label it into few co-ordinates that must stand in the circumference. To complete the circle we would change it upside down so that the circle. Students are being questioned to label circle where they are expected to respond from the interactive chart. After that students would take brief notes concerning what is the equation of a circle. They would also be questioned based on the previous lesson, what they know about the equation of a circle. Student must demonstrate their ability to reckon that circle has a radius, perimeter and the co-ordinates whose values rely on the understanding of the Cartesian plane. We would appreciate the Pythagoras theorem of the relationship between the hypotenuse. Basically learning Cartesian plane and Pythagoras theorem is crucial for understanding of the circle. The equation of the circle is derived from the distance formula whose points rely on the co-ordinate points. Then we rely on the Cartesian plan to derive points. The Cartesian plan is given by the intersection of the vertical and horizontal line. There is only one common point of intersection of Cartesian plane and that is called Origin O (o: o). in pure applied maths we often denote the origin to be a letter ‘O’ and it is located at the centre of the circle. We do not and it can never be possible to separate distance formula, mid-point formula and gradient to the equation of the circle. However that does not mean that lesson plan must simple talk about mid-point formula randomly. There has to be cognitive understanding and fully acknowledgeable reason behind the point of mentioning it. Children must develop conjectures and capabilities to reckon when, where and how to apply midpoint, distance formula and the gradient knowledge when dealing with the equation of a circle. In other hand it appear as if students are not aware that exam questions would not ask simple and straight forward questions. The lesson plan utilised the lesson observation for the previous exercise. In fact what was lacking is their ability to integrate algebra, geometry, exponential expressions and quadratic equations to the circle equations. In this lesson we are dealing with circle whose centre is located outside of the centre of the circle. Students are exposed to circle that is located at the centre of the circle whose equation is given by x2 + y2 = r2 . Generally students would only realise that to determine the equation of the circle you would only the centre and a radius. Their centre is origin which is not wrong, however there has to be transformation this time around. The change comes with the alteration of


centre of the circle to be situated somewhere outside the origin. Once again the distance formula is found to be useful in determining the equation of the circle whose point s outside of the equation. Development: The equation of a circle, centred off origin where (a: b) or (h: k) is called the centre of the circle. This equation will have a radius (r).

In other books they avoid indicating a circle because students often confuse the radius with the centre of the circle. Most books prefer to use dotted line excluding a line from centre to circumference. In the above drawing a radius is a distance between OD, where O is a centre of the circle. The centre of this circle is not centred at origin (0: 0). What do you think you can do to impress students to develop understanding of the equation of the circle whose centre is situated off the origin (0:0). NB radius is not the centre of the circle as most student


seem to confuse it. Rather the point that enables the radius to rotate around the central point from centre to the circumference of the circle is called the centre of a circle. From our previous lesson we noted that x 2 + y2 = r2. This formula is found through the use of the Pythagoras theorem, distance formula and understanding of a gradient.

In the previous lesson distance formula was used as follows: r = distance and x and y are a point at the circumference or a point where a radius meets the circumference (x: y). According to a distance formula; d2 = (x2-x1)2 – (y2-y1)2.

Where:

0: 0 ( x 1 : y 1)

and

x: y Âż x 2: y 2 There is certain hidden rule that students must discover and that is the fact that you must move from the circumference to the origin (from (x: y) to (0: 0). Therefore from the origin you will have (x1: y1) given by co-ordinates (0: 0) whereas the circumference would have (x2: y2) given by the co-ordinates (x: y). I

NB: It up to a teacher to see whether he can mark these previous lessons. (If not, then ignore it). But if he marks them, he has to make a drawing to illustrate this briefly. Questions are asked, students respond, where possible they stand up and pick the chalkboard to demonstrate their understanding. It is not a teacher centred, but learners participate in groups. It is not a teacher’s sole responsibility to recall circle types , students must also ask questions to challenge how critical a teacher demonstrate and thereby get their feedback. Similarly the equation of a circle centred off the origin is also given by a distance formula. In addition to that you must have a clue of working of algebraic expression, Pythagoras Theorem, gradient and the median and trigonometry. In any circle centred off origin the centre of the circle is given by (a: b) or (h: k) or what mode of denoting centre of a circle you may come across. Radius of a circle is measured can be calculated before one expresses the equation of the circle. This is derived from a distance formula.


Distance formula ( d): d2 = (x-a)2 + (y- b)2

or d2 = (x-h)2 + ( y-k)2

Do not forget to acknowledge the square root invers just in case you choose to make d the subject of a formula. d2 = r2 ∴

r2 = (x-a)2 + (y- b)2

, or r2 = (x-h)2 + ( y-k)2 students must know clear that ( a:b) is

centre of the circle whose central points are not at the origin (0: 0). The rule of expressing the equation of the circle centred off origin goes as follows: d2 = (x-a)2 + (y- b)2 Step 1: calculate the central point (a: b) depending which formula you have applied. Step 2: step 2 calculate the distance of a radius or simple calculate a radius distance.


Step 3: Eliminate the values of (x: y) but retain (a: b) and the value of r – the radius of a circle. Step 4 : Write down the equation of a circle , as required , do not leave answers in absurd form and where requested indicate a drawing of a circle. Remember, it is wise to write down an attempt to calculate equation even if you are not sure, to obtain marks from substitution. There are many ways of calculating at a radius distance, some can choose to calculate a diameter of a circle and half it. But the smartest way that is widely recommended by mathematician is to simple use a median formula (

x 1+ x 2 2

:

y 1+ y 2 ) to determine 2

the point of the centre of the circle , mostly the points of the centre of the circle centred off origin. Students need to acknowledge that there are very rare cases on which exam question is set to explore an essay and straight forward answer. Instead they will be required to show their applied maths skills, which need no maths background but simple understanding of development and application of a formula. That is why we often revise distance formula, algebra, midpoint formula and trigonometry whenever the function of a circle is being taught. Accordingly students would be advised to check the formula sheet at the back of their exam sheet. There is no need to panic about a formula that we are developing, it is part of revision exercise and means to let them master the circle function topic. NB: - To express an equation of the circle, following completion of your calculations, you must ensure to eliminate values of (x: y) but leave the values of the centre of a circle and radius. Planned questions:1) Where could you trace the emergence of the formula of a circle? 2) What is the difference between the equation of a circle centred off origin and the equation of a circle centred in the equation? 3) Mention at least three main important equations we have been discussing in dealing with circle equations. 4) What are three simple properties of a circle?


5) Given (x-2)2 + (y + 5)2 = 9, use the formula of the circle d2 = (x-h)2 + ( y-k)2 to determine the size of the radius and the central points of the circle. 6) What would be the equation of the circle centred off the origin , whose entre of a circle is (1: -4) and a radius =

√ 13

You may ignore the following: Challenges: Classroom Exercise: In the following circle suppose you were not given centre of the circle (-1: 2) Given: a circle with points a (-4: 3) and b (2: 1). Join the line ab to calculate the distance ab Let students determine points of the centre of the circle independently, but boost their knowledge where necessary. Y

X

Therefore determine: 1) 2) 3) 4)

The point of the centre of the circle The centre of the circle (a: b). Distance halfway from centre to the circumference ( radius) Express the equation of the circle into r2 = (x-a)2 + ( y-b)2 .


Assessment: Students would receive a feedback from a teacher’s memorandum informing them of how to calculate the equation of the circle centred off the origin. Corrections are going to be made and the exercise for application of the function of a circle are given from student textbook. Conclusion: The equation of a circle whose centre is not at the origin (0:0) is expressed as (a: b). We acknowledge the use of a distance formula, d2 = (x-a)2 + (y- b)2 for the development of an equation of a circle. For the love of this training students are advised to be able to separate difference between the equation of the circle centred at origin and the equation of the circle centred off origin. Generally the equation of a circle whose centre is at the origin (0: 0) is given by x2 + y2 = r2 . But the equation of the equation of the circle cantered off the origin ( a: b) is given by r2 = (x-a)2 + (y- b)2. This formula is very much same as a distance formula that is derived from the knowledge of how the Pythagoras Theorem works in a Cartesian plane as shown is a drawing of the above sections. Briefly the theorem of Pythagoras gave us an opened window to express the radius distance. In any circle centred at origin one can join the radius from circumference to the centre of a circle in order to calculate the size or distance covered by this point. Two points are joined, where the x-axis and y-axis converge towards their right angle. When that has happened we then argue that the radius is a hypotenuse. Then we apply the Pythagoras theorem which state that the square of the hypotenuse, the longest side of a triangle is equal to the square of the two sides standing against the right angle. But for the circle centred off origin we simple apply distance formula , but be careful do not confuse a centre of a circle from a radius as most student often appear to do that . Central point of a circle is given by (a: b) or (h: k) and not the radius itself. A radius is a distance from circumference of a circle to the centre of a circle. Students are advised to always look at the back of their exam book for their formula spread sheet. Many students forget to make use of those formula and expect to remember everything which does not work. Lastly: It is important to realise how reciprocals and inverse works when dealing with equations. This deeply lies from understanding of algebraic expression, trigonometry and exponents. Memorandum:


1) The emergence of a formula of a circle can be traced from Pythagoras theorem. The radius is produced from centre of the circle to the circumference. Then two points can be joined where a vertical line meet the apex of a radius at the circumference and the horizontal line meet the vertical line at the right angle so that they stand adjacent to the right angle. The radius becomes the hypotenuse as soon as the right angle is produced. Then the square of the hypotenuse is equal to the sum of the two squares of the sides that stand adjacent to the right angle and opposite to the hypotenuse. ( H 2 = D2 + V2) 2) The circle centres at origin has a centre of O (0; 0) whereas the equation of the circle centred off the origin has the centre (a: b). Therefore their equations also differ as the equation of the circle centred in the origin is given by x2 + y2 = r2. The equation of the circle centred off origin is given by r2 = (x−a) 2+( y −b)2

,

or r2 = (x-h)2 + ( y-k)2. Initially this formula is dated from a distance formula. 3) There are about five of them but the three most prominent where x 1+ x 2 y 1+ y 2 i) Mid-point formula: ( : ) . The midpoint formula is 2 2 used when two given point from the circle are given, where you need to determine the centre of the circle. You need to know the relationship between a radius and its circle, basic trigonometry, Pythagoras theorem and exponents ii)

to apply some geometric circle theories appropriately. Distance formula: d2 = (x-h)2 + ( y-k)2 , this is where a formula of a circle centred off origin is derived. Distance is similar to radius and that is why this become an equation of a circle centred off the origin. Remember (h:k) are central points of the circle. If you want to express your circle equation correctly you must begin by seeking to calculate the radius of the circle. The you must know how to solve for central points (a:b) or (h: k) where one of these points may form component of the radius or distance calculations. Once all values are known, you are encouraged to cancel out the values of x and y (x: y) but leave the values of a and b (a: b) into the equation. You are also required not to remove the actual value of a radius. Student often battle with

iii)

the concept if this was not explained in the class. Pythagoras Theorem: H2 = D2 + V2. Students need to consult grade 8 and 10 books for further illustration as they are expected to master the understanding of this theory briefly. The total number of square of the hypotenuse, the longest side of a right angled triangle is equal to the sum of two (2) number of


squares of the sides standing opposite to the longest side or adjacent to the right angle.

4) A circle is a circular drawing whose central points are limited to 3600. A circle’s circumference or rather the circumference of a circle is distance surround its point (perimeter). C = 2 π r. There are many ways of dealing with circle problem, sometimes we express its radius into diameter (d= 2r) and sometimes we deal with the radian. (

π 2

= 900). etc. . . Students are encouraged to consider revision of

trigonometric expressions, algebraic expressions and even exponents. e.g.)

1

√ Z = 12 n Z=

√ 1

1

12 n

(How to solve for z?)

usually we do not write one (1), so Z =

1

12 n

5) Given (x-2)2 + (y + 5)2 = 9, use the formula of the circle d2 = (x-h)2 + ( y-k)2 to determine the size of the radius and the central points of the circle. 6) (x-2)2 + (y + 5)2 = 9 (x-a)2 + (y- b)2 = r2 -a = -2, therefore a= 2 -b = 5, therefore b = -5 ∴ Centre of circle is (a: b) = (2: -5) r2 = 9, therefore r = 3, r ≠ -3 7) What would be the equation of the circle centred off the origin , whose entre of a circle is (1: -4) and a radius = √13 The equation of a circle is (x-a)2 + (y- b)2 = r2 (a: b) = ( 1: 4) , and r = √ 13 (x-1)2 + (y- 4)2 = (

13 √¿ ¿

2

(x-1)2 + (y- 4)2 =13. 8) i) The centre and radius is needed. The centre is a midpoint, so calculate the midpoint: x 1+ x 2 y 1+ y 2 Mid-pt = ( : ) . Where points a ( -4: 3) and b ( 2: 1) 2 2


Md-pt = (

−4+2 2

:

3+ 1 ) = (-1: 2) 2

(x1: y1)

&

b(x2: y2)

ii) Radius is a distance from centre to circumference I distance from centre is (-1: 2) to (2: 1) or is (-1: 2) to (-4: 3) radii. Students are encouraged to look for formula sheet at the back of their exam sheet. Distance formula d = √( x−a)2+( y−b)2 centre (a: b) = (-1: 2) vs any point (from circumference: e.g.: (2:1) = √( x−(−1))2+( y−2)2 = √ (x +1)2+( y−2)2 =

but point (x: y) = (2: 1)

√(2+1)2+(1−2)2

= √(3)2+(−1) 2 = √ 9+1 = √ 10 units Therefore r = √ 10 units Now, cancel off the x and y values but retain (a: b) and the values of the radius :( where… in the whole formula of calculation of a radius) (x+1)2 + (y- 22 = √ 10 2 (x+1)2 + (y- 22 = 102 The following is the figure to show how this problem was solved with the aid of Geogebra , Demos and plain GIS table. Geo-gebra is also available to use in the library with internet: students are advised to try it.


The end

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