C onics

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MTH-5105-1.indd 1
MTH-5105-1
onics
10/17/07 9:30:37 AM
MTH-5105-1
CONICS
Mathematics Project Coordinator: Jean-Paul Groleau
Author: Sylvie Amideneau
Updated Version: Line Régis; Christine Paris
Content Revision: Jean-Paul Groleau
Pedagogical Revision: Jean-Paul Groleau
Electronic Publishing: P.P.I. inc.
Translation: Claudia de Fulviis
Linguistic Revision: Johanne St-Martin
© Société de formation à distance des commissions scolaires du Québec
All rights for translation and adaptation, in whole or in part, reserved for all countries. Any
reproduction, by mechanical or electronic means, including microreproduction, is forbidden without the written permission of a duly authorized representative of the Société de
formation à distance des commissions scolaires du Québec (SOFAD).
Legal Deposit — 2007
Bibliothèque et Archives nationales du Québec
Bibliothèque et Archives Canada
ISBN 978-2-89493-316-9
November 2014
Answer Key
MTH-5105-1
Conics
TABLE OF CONTENTS
Introduction to the Program Flowchart ................................................. 0.4
Program Flowchart ................................................................................. 0.5
How to Use this Guide ............................................................................ 0.6
General Introduction ............................................................................... 0.9
Intermediate and Terminal Objectives of the Module .......................... 0.11
Diagnostic Test on the Prerequisites ..................................................... 0.17
Answer Key for the Diagnostic Test on the Prerequisites .................... 0.25
Analysis of the Diagnostic Test Results ................................................. 0.31
Information for Distance Education Students ....................................... 0.33
UNITS
1.
2.
3.
4.
5.
6.
7.
8.
General Form of the Equation of a Circle ............................................ 1.1
Graphing a Relation that Defines a Circle .......................................... 2.1
Equation of a Line Tangent to a Circle ................................................ 3.1
Standard Form of the Equation of a Parabola..................................... 4.1
Graphing a Relation Defining a Parabola ........................................... 5.1
Graphing an Ellipse Centred at the Origin ......................................... 6.1
Graphing a Hyperbola Centred at the Origin ..................................... 7.1
Equation or Inequality Associated with the Graph of a Conic
Section .................................................................................................... 8.1
9. Finding the Equation of a Conic Section Using the Description of
another Conic Section ........................................................................... 9.1
10. Equations and Geometric Loci ........................................................... 10.1
11. Solving Problems Involving Conics .................................................... 11.1
Final Review ........................................................................................ 12.1
Answer Key for the Final Review ....................................................... 12.9
Terminal Objectives ............................................................................ 12.14
Self-Evaluation Test............................................................................ 12.19
Answer Key for the Self-Evaluation Test .......................................... 12.33
Analysis of the Self-Evaluation Test Results .................................... 12.41
Final Evaluation............ ......................................................................12.42
Answer Key for the Exercises ............................................................. 12.43
Glossary ............................................................................................... 12.177
List of Symbols .................................................................................... 12.182
Bibliography ........................................................................................ 12.183
Review Activities ................................................................................. 13.1
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INTRODUCTION TO THE PROGRAM FLOWCHART
WELCOME TO THE WORLD OF MATHEMATICS
This mathematics program has been developed for adult students enrolled either
with Adult Education Services of school boards or in distance education. The
learning activities have been designed for individualized learning.
If you
encounter difficulties, do not hesitate to consult your teacher or to telephone the
resource person assigned to you. The following flowchart shows where this
module fits into the overall program. It allows you to see how far you have come
and how much further you still have to go to achieve your vocational objective.
There are three possible paths you can take, depending on your goal.
The first path, which consists of Modules MTH-3003-2 (MTH-314) and
MTH-4104-2 (MTH-416), leads to a Secondary School Vocational Diploma
(SSVD) and certain college-level programs for students who take MTH-4104-2.
The second path, consisting of Modules MTH-4109-1 (MTH-426), MTH-4111-2
(MTH-436) and MTH-5104-1 (MTH-514), leads to a Secondary School Diploma
(SSD), which gives you access to certain CEGEP programs that do not call for a
knowledge of advanced mathematics.
Lastly, the path consisting of Modules MTH-5109-1 (MTH-526) and MTH-5111-2
(MTH-536) will lead to CEGEP programs that require a thorough knowledge of
mathematics in addition to other abilities. Good luck!
If this is your first contact with the mathematics program, consult the flowchart
on the next page and then read the section “How to Use this Guide.” Otherwise,
go directly to the section entitled “General Introduction.” Enjoy your work!
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PROGRAM FLOWCHART
CEGEP
MTH-5112-1
MTH-5111-2
MTH-536
MTH-5104-1
MTH-5103-1
Introduction to Vectors
MTH-5109-1
Geometry IV
MTH-5108-2
Trigonometric Functions and Equations
MTH-5107-2
Exponential and Logarithmic Functions
and Equations
Optimization II
MTH-5106-1
Real Functions and Equations
Probability II
MTH-5105-1
Conics
MTH-5102-1
MTH-436
MTH-426
MTH-4110-1
MTH-216
MTH-116
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The Four Operations on
Algebraic Fractions
MTH-4109-1
Sets, Relations and Functions
Quadratic Functions
MTH-4107-1
Straight Lines II
MTH-4106-1
Factoring and Algebraic Functions
MTH-4105-1
Exponents and Radicals
MTH-4103-1
MTH-4102-1
MTH-4101-2
Complement and Synthesis I
MTH-4108-1
MTH-4104-2
MTH-314
Optimization I
MTH-4111-2
Trades
DVS
You ar e h er e
Statistics III
MTH-5101-1
MTH-416
Complement and Synthesis II
MTH-5110-1
MTH-526
MTH-514
Logic
Statistics II
Trigonometry I
Geometry III
Equations and Inequalities II
MTH-3003-2
Straight Lines I
MTH-3002-2
Geometry II
MTH-3001-2
The Four Operations on Polynomials
MTH-2008-2
Statistics and Probabilities I
MTH-2007-2
Geometry I
MTH-2006-2
Equations and Inequalities I
MTH-1007-2
Decimals and Percent
MTH-1006-2
The Four Operations on Fractions
MTH-1005-2
The Four Operations on Integers
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25 hours
= 1 credit
50 hours
= 2 credits
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Conics
HOW TO USE THIS GUIDE
Hi! My name is Monica and I have been
asked to tell you about this math module.
What’s your name?
Whether you are
registered at an
adult education
center or pursuing distance
education, ...
Now, the module you have in your
hands is divided into three
sections. The first section is...
I’m Andy.
... you have probably taken a
placement test which tells you
exactly which module you
should start with.
... the entry activity, which
contains the test on the
prerequisites.
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You’ll see that with this method, math is
a real breeze!
My results on the test
indicate that I should begin
with this module.
By carefully correcting this test using the
corresponding answer key, and recording your results on the analysis sheet ...
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... you can tell if you’re well enough
prepared to do all the activities in the
module.
And if I’m not, if I need a little
review before moving on, what
happens then?
Conics
In that case, before you start the
activities in the module, the results
analysis chart refers you to a review
activity near the end of the module.
Good!
In this way, I can be sure I
have all the prerequisites
for starting.
START
The starting line
shows where the
learning activities
begin.
Exactly! The second section
contains the learning activities. It’s
the main part of the module.
?
The little white question mark indicates the questions
for which answers are given in the text.
The target precedes the
objective to be met.
The memo pad signals a brief reminder of
concepts which you have already studied.
?
Look closely at the box to
the right. It explains the
symbols used to identify the
various activities.
The boldface question mark
indicates practice exercises
which allow you to try out what
you have just learned.
The calculator symbol reminds you that
you will need to use your calculator.
?
The sheaf of wheat indicates a review designed to
reinforce what you have just learned. A row of
sheaves near the end of the module indicates the
final review, which helps you to interrelate all the
learning activities in the module.
FINISH
Lastly, the finish line indicates
that it is time to go on to the self-evaluation
test to verify how well you have understood
the learning activities.
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There are also many fun things
in this module. For example,
when you see the drawing of a
sage, it introduces a “Did you
know that...”
It’s the same for the “math whiz”
pages, which are designed especially for those who love math.
For example. words in boldface italics appear in the
glossary at the end of the
module...
A “Did you know that...”?
Yes, for example, short tidbits
on the history of mathematics
and fun puzzles. They are interesting and relieve tension at
the same time.
Conics
Must I memorize what the sage says?
No, it’s not part of the learning activity. It’s just there to
give you a breather.
And the whole module has
been arranged to make
learning easier.
They are so stimulating that
even if you don’t have to do
them, you’ll still want to.
... statements in boxes are important
points to remember, like definitions, formulas and rules. I’m telling you, the format makes everything much easier.
The third section contains the final review, which interrelates the different
parts of the module.
Great!
There is also a self-evaluation
test and answer key. They tell
you if you’re ready for the final
evaluation.
Thanks, Monica, you’ve been a
big help.
I’m glad! Now,
I’ve got to run.
See you!
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Later ...
This is great! I never thought that I would
like mathematics as much as this!
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GENERAL INTRODUCTION
CONIC SECTIONS
This module deals with conics, that is, figures obtained by cutting a cone with a
plane at different angles. You are already familiar with the parabola as you
studied it in Module MTH-4108-1. In this module, we will look at the parabola
again. This time, however, we will consider it as a set of points equidistant from
a fixed point called the “focus” and a straight line called the “directrix.” The focus
of the parabolas covered in this module is any point in the plane and the directrix
is a vertical or horizontal line.
We will also look at the circle, which is a set of points equidistant from a fixed
point called the “centre” of the circle. We will learn how to find the equation of
a circle given the coordinates of its centre and the measurement of its radius and
the equation of a line tangent to a circle at a given point.
Lastly, we will study the ellipse and the hyperbola. An ellipse is a set of points
such that the sum of the distances from two fixed points, called the foci, is
constant. The hyperbola is a set of points such that the difference of the distances
from two fixed points, called the foci, is constant. For these two conic sections,
we will look at cases where the centre of the figure coincides with the origin of
the Cartesian coordinates, or point (0, 0).
We will see that each of these conic sections defines three regions in the plane.
Given equations and inequalities, we will graph the corresponding regions of the
plane and find the domain and the range associated with each relation.
At the end of the module, we will do the reverse. We will find the equation or
inequality associated with the graph of a given conic relation.
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We will then look at how to find the equation of a conic section from another conic
section or from its geometric locus in order to be able to solve problems involving
conics.
The world of conics might seem strange to you at first; however, with order and
method, you will be able to understand it.
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INTERMEDIATE AND TERMINAL OBJECTIVES
OF THE MODULE
Module MTH-5105-1 consists of 11 units and requires 25 hours of study,
distributed as shown below. The terminal objectives appear in boldface.
Objectives
Number of Hours*
% (Evaluation)
1
2
5%
2
2
5%
3
2
5%
4
2
5%
5
2
5%
6
2
5%
7
2
5%
8
2
5%
9
3
20%
10
2
5%
11
3
20%
* One hour is allotted for the final evaluation.
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Conics
1. General form of the equation of a circle
Find
the
general
form
of
the
equation
of
a
circle
(i.e. x2 + y2 + Dx + Ey + F = 0), given its centre (h, k) and its radius r.
Conversely, find the centre (h, k) and the radius r of a circle, given its
equation in general form. The parameters D, E, F, h, k and r are
rational numbers and are usually integers.
2. Graphing a relation that defines a circle
Graph the region determined by a relation defining a circle. Find the
domain and the range of this relation and indicate them using
interval or set-builder notation.
The relation can be written in one of the following forms:
• x2 + y2 + Dx + Ey + F = 0 or (x – h)2 + (y – k)2 = r2
• x2 + y2 + Dx + Ey + F < 0 or (x – h)2 + (y – k)2 < r2
• x2 + y2 + Dx + Ey + F ≤ 0 or (x – h)2 + (y – k)2 ≤ r2
• x2 + y2 + Dx + Ey + F > 0 or (x – h)2 + (y – k)2 > r2
• x2 + y2 + Dx + Ey + F ≥ 0 or (x – h)2 + (y – k)2 ≥ r2
The parameters D, E, F, h, k and r are rational numbers. Indicate the centre
of the circle and its radius clearly on the graph.
3. Equation of a line tangent to a circle
Find the equation of a line tangent to a circle, given the point of
tangency (x1, y1) and the equation of the circle in standard form [i.e.
(x – h)2 + (y – k)2 = r2] or in general form [i.e. x2 + y2 + Dx + Ey + F = 0].
The parameters x1, y1, h, k, r, D, E and F are integers.
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4. Standard form of the equation of a parabola
Find the standard form of the equation of a parabola
[i.e. (y – k)2 = ± 4a(x – h) or (x – h)2 = ± 4a(y – k)], given its vertex (h, k)
and its focus (x1, y1). The parameters a, h, k, x1 and y1 are integers and
a ≠ 0.
Find and graph the equation of a parabola obtained after a given parabola
with vertex at the origin (i.e. y2 = ± 4ax or x2 = ± 4ay) has been translated so
that its vertex becomes (h, k). The contants h and k are integers, and a is a
natural number other than 0.
5. Graphing a relation defining a parabola
Graph a region determined by a relation defining a parabola. Find
the domain and the range of this relation and indicate them using
interval or set-builder notation.
The relation can be written in one of the following forms:
• (y – k)2 = ± 4a(x – h) or (x – h)2 = ± 4a(y – k)
• (y – k)2 < ± 4a(x – h) or (x – h)2 < ± 4a(y – k)
• (y – k)2 ≤ ± 4a(x – h) or (x – h)2 ≤ ± 4a(y – k)
• (y – k)2 > ± 4a(x – h) or (x – h)2 > ± 4a(y – k)
• (y – k)2 ≥ ± 4a(x – h) or (x – h)2 ≥ ± 4a(y – k)
The parameter a is a rational number other than 0. The parameters h and k
are also rational numbers but are usually integers. Clearly indicate the
following on the graph: the vertex, the focus, the axis of symmetry and the
directrix of the parabola.
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6. Graphing an ellipse centred at the origin
Graph a region determined by a relation defining an ellipse centred
at the origin. Indicate the axes and the two foci of the ellipse clearly
on the graph. Find the domain and the range of this relation and
indicate them using interval or set-builder notation.
The relation can be written in one of the following forms:
•
2
x2 + y = 1
2
2
a
b
•
2
x2 + y < 1
2
2
a
b
•
2
x2 + y ≤ 1
2
2
a
b
•
2
x2 + y > 1
2
2
a
b
•
2
x2 + y ≥ 1
2
a
b2
The parameters a and b are integers other than 0.
7. Graphing a hyperbola centred at the origin
Graph a region determined by a relation defining a hyperbola
centred at the origin. Indicate the asymptotes, vertices and foci of
the hyperbola clearly on the graph. Find the domain and the range
of this relation and indicate them using interval or set-builder
notation.
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The relation can be written in one of the following forms:
•
2
x2 – y = 1
2
2
a
b
•
y2 x2
–
= 1 or
b2 a2
2
x2 – y = –1
2
2
a
b
•
2
x2 – y < 1
2
2
a
b
•
y2 x2
–
< 1 or
b2 a2
2
x2 – y > –1
2
2
a
b
•
2
x2 – y ≤ 1
2
2
a
b
•
y2 x2
– 2 ≤ 1 or
2
a
b
2
x2 – y ≥ –1
2
2
a
b
•
2
x2 – y > 1
2
2
a
b
•
y2 x2
– 2 > 1 or
2
a
b
2
x2 – y < –1
2
2
a
b
•
x2 – y2 ≥ 1
a2 b2
•
y2 x2
– 2 ≥ 1 or
2
a
b
x2 – y2 ≤ –1
a2 b2
The parameters a and b are integers other than 0.
8. Equation or inequality associated with the graph of a conic section
Find the equation or inequality associated with the graph of one of
the following conic sections: a circle, a parabola, an ellipse centred
at the origin or a hyperbola centred at the origin. The graph may or
may not contain a shaded region, depending on the situation. The
distinctive features of each curve are clearly indicated on the given
graph (i.e. the radius, the centre, one or more vertices, one or more
foci, one or more axes of symmetry, the directrix, the asymptotes, as
the case may be). The equation or inequality must be written in
standard form.
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9. Finding the equation of a conic section using the description of
another conic section
Find the equation of a conic section using the description of another
conic section. The equation or certain features of this other conic
section are known. These conics and their respective features can
be any of the following: a circle (centre and radius); a parabola
(vertex, focus and directrix); an ellipse and a hyperbola centred at
the origin (vertices, foci, equations of the asymptotes). The features
may be described indirectly.
10. Equations and geometric loci
Given their definition as a locus, find the equation of one of the
following conics: a circle, a parabola, an ellipse centred at the origin
and a hyperbola centred at the origin.
11. Solving problems involving conics
Solve problems that involve applying concepts related to the following conics: a circle, a parabola, an ellipse centred at the origin or a
hyperbola centred at the origin. The solution may require finding
an equation describing a relation, drawing a graph, determining
the coordinates of certain points and calculating the distance
between certain points. The equations should be given in standard
form only, with the exception of those associated with a circle,
which can be defined in general form.
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DIAGNOSTIC TEST ON THE PREREQUISITES
Instructions
1. Answer as many questions as you can.
2. You may use a calculator.
3. Write your answers on the test paper.
4. Do not waste any time. If you cannot answer a question, go on
to the next one immediately.
5. When you have answered as many questions as you can, correct
your answers using the answer key which follows the diagnostic
test.
6. To be considered correct, your answers must be identical to
those in the key. In addition, the various steps in your solution
should be equivalent to those shown in the answer key.
7. Transcribe your results on the chart which follows the answer
key. This chart gives an analysis of the diagnostic test results.
8. Do only the review activities that apply to each of your incorrect
answers.
9. If all your answers are correct, you may begin working on this
module.
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1. What is the slope of the line passing through points (5, – 8) and (– 2, 7)?
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2. Find the equations of the lines that satisfy the following conditions. Check
your answers.
a) A line passing through point (– 3, 2) and whose slope is 1 .
2
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b) A line passing through point (– 3, 2) and whose slope is – 3 .
7
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c) A horizontal line passing through point (2, 7).
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d) A vertical line passing through point (– 2, – 3).
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3. What is the slope of a line perpendicular to a line whose slope is 4 ?
5
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4. Find the distance between the following points.
a) (3, 7) and (3, – 5)
b) (– 4, 2) and (– 7, 2)
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c) (0, 3) and (– 4, 0)
d) (7, – 2) and (– 2, 5)
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5. Factor the following polynomials and check your answers.
a) x2 – 16x + 64
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b) 4x2 + 28x + 49
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6. Solve the following equations by showing all the steps in your solution and
check your answers.
a) x2 – 36 = 0
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b) 2x2 = 250
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c) x2 + 10x – 11 = 0
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d) x2 – 5x + 6 = 0
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ANSWER KEY FOR THE DIAGNOSTIC TEST
ON THE PREREQUISITES
1. Let (x1, y1) = (5, – 8) and (x2, y2) = (– 2, 7).
y –y
m = x2 – x 1
2
1
m=
7 – (– 8) 7 + 8
=
= – 15
–7
7
–2 – 5
2. a) Let (x1, y1) = (– 3, 2) and m = 1 .
2
y–y
m = x – x1
1
Check
Let (x, y) = (– 3, 2).
1 = y–2
2
x – (– 3)
x – 2y + 7= 0
1 = y–2
2
x+3
1(x + 3) = 2( y – 2)
– 3 – 2(2) + 7 = 0
–3 – 4 + 7 = 0
–7 + 7 = 0
x + 3= 2y – 4
0=0
x – 2y + 3 + 4= 0
x – 2y + 7 = 0
b) Let (x1, y1) = (– 3, 2) and m = – 3 .
7
y – y1
m= x–x
1
Let (x, y) = (– 3, 2).
y–2
–3 =
7
x – (– 3)
3x + 7y – 5 = 0
y–2
–3 =
7
x+3
– 3(x + 3) = 7( y – 2)
3(– 3) + 7(2) – 5 = 0
– 9 + 14 – 5 = 0
0=0
– 3x – 9 = 7y – 14
– 3x – 7y – 9 + 14 = 0
– 3x – 7y + 5 = 0
× (– 1)
3x + 7y – 5 = 0
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c) The equation of a horizontal line is of the form y = b.
Since b = 7, then y = 7.
Check
Let (x, y) = (2, 7).
y=7
y = 0x + 7
7 = 0(2) + 7
7=0+7
7=7
d) The equation of a vertical line is of the form x = a.
Since a = – 2, then x = – 2.
Check
Let (x, y) = (– 2, – 3).
x = –2
x = 0y – 2
– 2 = 0(– 3) – 2
–2 = 0 – 2
–2 = –2
3. Two lines are perpendicular if and only if the product of their slopes
equals – 1.
Hence, m1 = – m1 or m2 = – m1 .
2
1
If m1 = 4 , then m2 = – 1 = –1 × 5 = – 5 .
5
4
4
4
5
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4. a) As the x-coordinates are the b) As the y-coordinates are the same,
same, you need only find the
you need only find the difference
difference between the y-coordi-
between the x-coordinates.
nates.
d = ⏐y2 – y1⏐
d = ⏐x2 – x1⏐
d = ⏐– 5 – 7⏐
d = ⏐– 7 – (– 4)⏐
d = ⏐– 12⏐
d = ⏐– 7 + 4⏐
d= 12
d = ⏐– 3⏐
d=3
c) Let (x1, y1) = (0, 3),
d) Let (x1, y1) = (7, – 2),
(x2, y2) = (– 4, 0).
d=
x2 – x1
d=
–4 – 0
d=
–4
2
2
2
(x2, y2) = (– 2, 5).
+ y2 – y1
+ 0–3
+ –3
2
2
2
d=
x 2 – x1
d=
–2 – 7
d=
–9
2
2
2
+ y2 – y1
+ 5 – (– 2)
+ 7
2
d = 16 + 9 = 25
d = 81 + 49 = 130
d=5
d 11.4
5. a) It is a trinomial of the form x2 + bx + c.
x2 – 16x + 64
Check
x2 – 8x – 8x + 64
x(x – 8) – 8(x – 8)
(x – 8)(x – 8)
(x – 8)(x – 8) or (x – 8)2
(x – 8)x + (x – 8)(– 8)
x2 – 8x – 8x + 64
x2 – 16 x + 64
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b) It is a trinomial of the form ax2 + bx + c.
4x2 + 28x + 49
Check
4x2 + 14x + 14x + 49
2x(2x + 7) + 7(2x + 7)
(2x + 7)(2x + 7)
(2x + 7)(2x + 7) or (2x + 7)2
(2x + 7)2x + (2x + 7)7
4x2 + 14x + 14x + 49
4x2 + 28x + 49
6. a) Difference of two squares.
x2 – 36 = 0
or
x2 – 36 = 0
x2 = 36
(x – 6)(x + 6) = 0, then
x = ± 36
x–6=0
or
x=6
x+6=0
x = ±6
x = –6
Check
• If x = 6, then
or
If x = – 6, then
x2 – 36 = 0
x2 – 36 = 0
(6)2 – 36 = 0
(– 6)2 – 36 = 0
36 – 36 = 0
36 – 36 = 0
0=0
True
0 = 0 True
Answer: x = 6 or x = – 6.
N.B. You can also write x = ± 6.
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b) 2x2 = 250
or
Conics
Using the quadratic formula:
x2 = 125
x = ± 125
2x2 = 250
x = ±11.18
2x2 – 250 = 0
a = 2, b = 0 and c = – 250.
2
x = – b ± b – 4ac
2a
x=
0 ± 0 2 – 4(2)(– 250)
2(2)
x = 0 ± 2 000 = x = ±44.72 = ±11.18
4
4
Check
• If x = 11.18, then
If x = – 11.18, then
2x2 = 250
2x2 = 250
2(11.18)2 = 250
2(– 11.18)2 = 250
2(124.99) = 250
2(124.99) = 250
249.98 ≈ 250
249.98 ≈ 250
True
Answer: x = 11.18 or x = – 11.18.
N.B. You can also write x = ±11.18.
c) Trinomial of the form x2 + bx + c.
x2 + 10x – 11 = 0
x2 + 11x – 1x – 11 = 0
x(x + 11) – 1(x + 11) = 0
(x – 1)(x + 11) = 0, then x – 1 = 0
x=1
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or
x + 11 = 0
x = – 11
True
1
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3
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Conics
Check
• If x = – 11, then
or
If x = 1, then
x2 + 10x – 11 = 0
x2 + 10x – 11 = 0
(– 11)2 + 10(– 11) – 11 = 0
(1)2 + 10(1) – 11 = 0
121 – 110 – 11 = 0
1 + 10 – 11 = 0
0 = 0 True
0 = 0 True
Answer: x = – 11 or x = 1.
d) Trinomial of the form x2 + bx + c.
x2 – 5x + 6 = 0
x2 – 2x – 3x + 6 = 0
x(x – 2) – 3(x – 2) = 0
(x – 2)(x – 3) = 0, then x – 2 = 0 or x – 3 = 0
x=2
x=3
Check
• If x = 2, then
If x = 3, then
x2 – 5x + 6 = 0
x2 – 5x + 6 = 0
(2)2 – 5(2) + 6 = 0
(3)2 – 5(3) + 6 = 0
4 – 10 + 6 = 0
9 – 15 + 6 = 0
0 = 0 True
0 = 0 True
Answer: x = 2 or x = 3.
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ANALYSIS OF THE DIAGNOSTIC
TEST RESULTS
Questions
Answers
Page
1.
13.2.1
13.11
Unit 3
2. a)
13.2.3
13.18
Unit 4
b)
13.2.3
13.18
Unit 4
c)
13.2.4
13.22
Unit 5
13.2.4
13.22
Unit 5
3. a)
13.2.4
13.15
Unit 4
4. a)
13.1.1
13.4
Unit 1
b)
13.1.1
13.4
Unit 1
c)
13.1.2
13.8
Unit 6
d)
13.3.1
13.8
Unit 6
5. a)
13.3.1
12.24
Unit 2
b)
13.3.1
12.24
Unit 2
6. a)
13.3.2
13.37
Unit 5
b)
13.3.2
13.37
Unit 5
c)
13.3.2
13.37
Unit 7
d)
13.3.2
13.37
Unit 7
ad)d))
Incorrect
Before Going on to
Section
d)
Correct
Review
• If all your answers are correct, you may begin working on this module.
• For each incorrect answer, find the related section listed in the Review
column. Do the review activities for that section before beginning the unit
listed in the right-hand column under the heading Before Going on to.
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INFORMATION FOR DISTANCE
EDUCATION STUDENTS
You now have the learning material for MTH-5105-1 and the relevant homework
assignments. Enclosed with this package is a letter of introduction from your
tutor, indicating the various ways in which you can communicate with him or her
(e.g. by letter or telephone), as well as the times when he or she is available. Your
tutor will correct your work and help you with your studies. Do not hesitate to
make use of his or her services if you have any questions.
DEVELOPING EFFECTIVE STUDY HABITS
Learning by correspondence is a process which offers considerable flexibility, but
which also requires active involvement on your part. It demands regular study
and sustained effort. Efficient study habits will simplify your task. To ensure
effective and continuous progress in your studies, it is strongly recommended
that you:
• draw up a study timetable that takes your work habits into account and is
compatible with your leisure and other activities;
• develop a habit of regular and concentrated study.
The following guidelines concerning theory, examples, exercises and assignments are designed to help you succeed in this mathematics course.
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Theory
To make sure you grasp the theoretical concepts thoroughly:
1. Read the lesson carefully and underline the important points.
2. Memorize the definitions, formulas and procedures used to solve a given
problem; this will make the lesson much easier to understand.
3. At the end of the assignment, make a note of any points that you do not
understand using the sheets provided for this purpose. Your tutor will then
be able to give you pertinent explanations.
4. Try to continue studying even if you run into a problem. However, if a major
difficulty hinders your progress, contact your tutor before handing in your
assignment, using the procedures outlined in the letter of introduction.
Examples
The examples given throughout the course are applications of the theory you are
studying. They illustrate the steps involved in doing the exercises. Carefully
study the solutions given in the examples and redo the examples yourself before
starting the exercises.
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Exercises
The exercises in each unit are generally modeled on the examples provided. Here
are a few suggestions to help you complete these exercises.
1. Write up your solutions, using the examples in the unit as models. It is
important not to refer to the answer key found on the coloured pages at the
back of the module until you have completed the exercises.
2. Compare your solutions with those in the answer key only after having done
all the exercises. Careful! Examine the steps in your solutions carefully,
even if your answers are correct.
3. If you find a mistake in your answer or solution, review the concepts that you
did not understand, as well as the pertinent examples. Then redo the
exercise.
4. Make sure you have successfully completed all the exercises in a unit before
moving on to the next one.
Homework Assignments
Module MTH-5105-1 comprises three homework assignments. The first page of
each assignment indicates the units to which the questions refer. The assignments are designed to evaluate how well you have understood the material
studied. They also provide a means of communicating with your tutor.
When you have understood the material and have successfully completed the
pertinent exercises, do the corresponding assignment right away. Here are a few
suggestions:
1. Do a rough draft first, and then, if necessary, revise your solutions before
writing out a clean copy of your answer.
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2. Copy out your final answers or solutions in the blank spaces of the document
to be sent to your tutor. It is best to use a pencil.
3. Include a clear and detailed solution with the answer if the problem involves
several steps.
4. Mail only one homework assignment at a time. After correcting the assignment, your tutor will return it to you.
In the section “Student’s Questions,” write any questions which you wish to have
answered by your tutor. He or she will give you advice and guide you in your
studies, if necessary.
In this course
Homework Assignment 1 is based on units 1 to 7.
Homework Assignment 2 is based on units 8 to 11.
Homework Assignment 3 is based on units 1 to 11.
CERTIFICATION
When you have completed all your work, and provided you have maintained an
average of at least 60%, you will be eligible to write the examination for this
course.
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START
UNIT 1
GENERAL FORM OF THE EQUATION OF A
CIRCLE
1.1
SETTING THE CONTEXT
Both Feet in the Sand
Janie is vacationing by the seaside. The sand on the beach where she is walking
is packed down hard and is kept moist by the incoming waves. Janie drives a
piece of wood into the sand to which she ties a string 20 cm long. To the end of
this string she ties a nail. She is having fun observing the line that the nail traces
in the sand when she pulls the string out as far as possible and goes all around
the piece of wood. Nearby, her brother Gilbert is doing the same thing with a
string 30 cm long. Have Janie and Gilbert traced a similar drawing in the sand?
What is this drawing called?
To achieve the objective of this unit, you should be able to find the
general form of the equation of a circle given the coordinates of its
centre (h, k) and its radius r.
You may have guessed that both Janie and Gilbert traced a circle. Gilbert’s circle
has a radius of 30 cm; therefore, it is larger than Janie’s circle, which has a radius
of only 20 cm.
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y
Let’s now look at the equation of a circle
with radius r and whose centre is situated at point (0, 0), which is the origin of
a Cartesian plane. First, remember that
(x, y)
r
all the points on a circle are equidistant
•
1
from a fixed point called the “centre” of
x
1
the circle and that this distance corresponds to the radius of the circle. Each
point (x, y) on a circle centred at the
origin therefore lies a fixed distance r
from point (0, 0).
Fig. 1.1
Circle with radius r centred
at the origin
Before finding the equation of this circle, let’s review the formula for finding the
distance between two points.
The distance between points P1(x1, y1 ) and P2(x2, y2 ) is obtained
using the formula
x2 – x1
2
+ y2 – y1
2
.
Let’s now find the equation of the circle in Figure 1.1.
Let d = r, (x1, y1) = (0, 0) and (x2, y2) = (x, y), which could be any point on the circle.
We can therefore write:
d=
x 2 – x1
r=
x–0
r=
x2 + y2
1.2
2
2
+ y2 – y1
+ y–0
2
2
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Square both sides of the equation.
x2 + y2
(r)2 =
2
r 2 = x2 + y2
The result is x2 + y2 = r2.
This equation is the general form of the equation of a circle with radius r and
centred at the origin.
The general form of the equation of a circle with radius r
and centred at the origin is:
x 2 + y2 = r 2
Exercise 1.1
1. Find the radius of each circle whose equation is given below.
a) x2 + y2 = 64 .............................
b) x2 + y2 = 49 ........................
c) x2 + y2 = 100 ..........................
d) x2 + y2 = 81 ........................
2. For each equation given below, write the equation of the circle centred at the
origin.
a) r = 1 ........................................
b) r = 5...................................
c) r = 13 ......................................
d) r = 15.................................
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Of course, a circle is not always centred at the origin. It can be moved in a given
direction in the plane; this is called a translation.
A translation is a transformation where each point of a figure
slides the same distance and direction to a new position in the
plane.
Given circle x2 + y2 = 36. It has centre (0, 0) and radius 6 units. Two translations
are applied to this circle. The first shifts it 5 units to the right and the second
shifts it 4 units upwards.
After the first translation, the circle's centre is (5,0).
y
1
•
1
Fig. 1.2
(5, 0)
x
Circle x2 + y2 = 36 is translated
5 units to the right
1.4
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After the second translation, the circle's centre is (5,4).
y
• (5, 4)
1
•
1
Fig. 1.3
(5, 0)
x
Circle x2 + y2 = 36 is then translated
4 units upwards
The same result could have been obtained with a single translation of (5, 4) which
moves the circle 5 units to the right and 4 units upwards.
Since a translation does not modify the figure, the image resulting from this
translation of (5, 4) units is still a circle with radius 6 units. However, the centre
of the circle is now (5, 4). To find the equation of the image obtained by applying
this translation to the original circle, remember that all points (x, y) on the circle
are 6 units from the centre (5, 4). Apply the formula for calculating the distance.
d=
(x 2 – x 1) 2 + ( y 2 – y 1) 2 where d = 6, (x1, y1) = (5, 4) and x2, y2 = (x, y)
6=
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(x – 5) + ( y – 4)
1.5
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Conics
Square both sides of the equation.
2
2
2
(x – 5) + ( y – 4)
(6) =
2
36 = (x – 5)2 + (y – 4)2
or (x – 5)2 + (y – 4)2 = 36
y
(x, y)
When a translation of (h, k) units is
r
applied to a circle with radius r, the
•
(h, k)
centre becomes (h, k) and the radius r
1
remains the same. All points (x, y) on the
x
1
circle are still located r units from point
(h, k).
Fig. 1.4
Circle with radius r and
centre (h, k)
We can therefore apply the formula for finding the distance between two points.
d=
(x 2 – x 1) 2 + ( y 2 – y 1) 2 where d = r, (x1, y1) = (h, k) and (x2, y2) = (x, y).
r=
(x – h) 2 + ( y – k) 2
r2 =
(x – h)2 + (y – k)2
or (x – h)2 + (y – k)2 = r2
This equation is called the standard form of the equation of a circle.
The standard form of the equation of a circle with centre
(h, k) and radius r is:
(x – h)2 + (y – k)2 = r2
1.6
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Example 1
If Janie’s circle is translated (9, 6) units, find the equation of the new circle
and graph it.
Since the circle has been translated (9, 6) units, the centre is now (9, 6). The
equation is therefore (x – h)2 + (y – k)2 = r2, where (h, k) = (9, 6) and r = 20.
∴ (x – 9)2 + (y – 6)2 = 400
N.B. To draw this circle, place the sharp point of the compass at point
(9, 6) and adjust the other compass arm so that the distance between each tip
is 20 units.
y
20
•
(9, 6)
3
3
Fig. 1.5
x
Graph of Janie's circle after a translation of (9, 6) units
has been applied to it
?
If Gilbert’s circle is translated (– 5, 5) units, find the equation of the new
circle and graph it.
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y
5
5
x
Fig. 1.6 Cartesian plane
Since the circle has been translated (– 5, 5) units, the centre of the circle is now
(– 5, 5).
Therefore, the equation is:
(x – h)2 + (y – k)2 = r2, where (h, k) = (– 5, 5) and r = 30.
(x – (– 5))2 + (y – 5)2 = 302
(x + 5)2 + (y – 5)2 = 900
1.8
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y
•5
x
5
Fig. 1.7
Circle (x + 5)2 + ( y – 5)2 = 900
y
?
Given the circle on the right
which is centred at the origin,
graph the new circle obtained by
applying a translation of (3, 2)
1
units. What are the equations of
1
the circle shown on the right and
of the circle resulting from the
translation?
Fig. 1.8
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1.9
Circle centred at the origin
with radius 3 units
x
1
Answer Key
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Give:
a) the equation of the circle centred at the origin;
.........................................................
b) the equation of the circle resulting from the translation.
.........................................................
Solution
y
The equation of the circle centred at the
origin is x2 + y2 = 9, since its radius is
•(3, 2)
1
3 units. The equation of the new circle is
x
1
(x – 3)2 + (y – 2)2 = 9.
Fig. 1.9 Circles whose equations are
x2 + y2 = 9 and (x – 3)2 + (y – 2)2 = 9
Did you know that...
... the radius of a circle is not always a whole number? It can
be an irrational number. Thus, for the circles whose equations are x2 + y2 = 2 and x2 + y2 = 3, the radii are
2 and
3 units respectively.
1.10
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y
(–1, 1)
•
y
(1, 1)
1
•
1
(–1, –1)
•
Conics
1
x
1
• (1, –1)
Fig. 1.10 Circle x2 + y2 = 2
Fig. 1.11 Circle x2 + y2 = 3
The equation x2 + y2 = 2 is a Diophantine equation as it has four solutions
which are whole numbers (represented by points). However, the equation
x2 + y2 = 3 is not a Diophantine equation as none of its solutions are whole
numbers. In fact, none of its points (x, y) consist of two whole-number
coordinates.
This type of equation was named after Diophantus, a Greek mathematician who lived in the third century A.D.
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Exercise 1.2
1. Find the equation of the circle obtained by applying the required translation
and graph it.
y
a) Translation of (0, 4) units applied
to the circle x2 + y2 = 36.
Equation:
2
......................................................
2
x
2
x
y
b) Translation of (– 3, 0) units applied to the circle x2 + y2 = 25.
Equation:
2
......................................................
1.12
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y
c) Translation of (3, – 2) units applied to the circle x2 + y2 = 9.
Equation:
1
......................................................
1
x
2. On each set of axes below, graph the circle obtained by applying the required
translation and give its equation.
y
a) Translation of (– 4, 0) units.
Equation:
......................................................
1
1
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y
b) Translation of (4, 3) units.
Equation:
......................................................
1
1
x
1
x
y
c) Translation of (6, – 5) units.
Equation:
......................................................
1
Now that you are familiar with the standard form of a circle with radius r and
centred at point (h, k), let's look at how this equation can be expressed in another
form.
1.14
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Let's look again at our circle centred at the origin and with a radius of 6 units.
We applied a translation of (5, 4) units to it and the radius remained the same,
i.e. 6 units. We found that the standard equation of this circle is
(x – 5)2 + (y – 4)2 = 36.
We can transform this equation by finding the squares and simplifying like terms.
(x – 5)2 + (y – 4)2 = 36
x2 – 10x + 25 + y2 – 8y + 16 = 36
x2 + y2 – 10x – 8y + 25 + 16 – 36 = 0
x2 + y2 – 10x – 8y + 5 = 0
This last equation is called the "general form of the equation of the circle" with
centre (5, 4) and radius 6 units.
This equation is of the form x2 + y2 + Dx + Ey + F = 0, where D = – 10, E = – 8 and
F = 5.
Example 2
Given a circle with centre 1, 1
2
and radius 2 units.
2
The standard equation of this circle is (x – 1) + y – 1
2
2
= 22.
Find the squares and simplify like terms.
x 2 – 2x + 1 + y 2 – y + 1 = 4
4
x 2 + y 2 – 2x – y + 4 + 1 – 16 = 0
4 4
4
x 2 + y 2 – 2x – y – 11 = 0
4
This equation is of the form x2 + y2 + Dx + Ey + F = 0, where D = – 2,
E = – 1 and F = – 11 .
4
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The general form of the equation of a circle is:
x2 + y2 + Dx + Ey + F = 0
To find the general form of the equation of a circle with
centre (h, k) and radius r:
1. write the equation of the circle in its standard form:
(x – h)2 + (y – k)2 = r2;
2. find the squares;
3. simplify like terms and arrange them so as to obtain the
equation of the form x2 + y2 + Dx + Ey + F = 0.
Let's now apply this method to the following example.
Example 3
Find the general form of the equation of the circle with centre (7, – 2) and
radius 4 units.
(x – 7)2 + (y + 2)2 = 42
x2 – 14x + 49 + y2 + 4y + 4 = 16
x2 + y2 – 14x + 4y + 37 = 0
?
From among the following second-degree equations in two variables, find
the four equations that represent circles.
➀ 5x2 + 6y2 = 45
➁ 3x2 + 3y2 – 9x + 12y – 27 = 0
➂ x2 + y2 – 8x + 6y – 40 = 0
➃ 6x2 + 6y2 – 36x + 48y – 98 = 0
➄ x2 – y2 – 7x – 26y + 64 = 0
➅ x2 – 4x + 18y + 12 = 0
➆ 9x2 + 9y2 – 81x + 108y + 100 = 0
➇ y2 + x2 – 4y + 80 = 0
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Solution
Your answer should be equations nos. ➁, ➂, ➃ and ➆. Equation no. ➂ is corrrectly
written in the general form x2 + y2 + Dx + Ey + F = 0, where D = –8, E = 6 and
F = 40. Equations ➁, ➃ and ➆ are written in the general form but we need only
divide the coefficients by 3 (for equation ➁), by 6 (for equation ➃) and by 9 (for
equation ➆) to obtain the general form x2 + y2 + Dx + Ey + F = 0.
N.B. In all the other equations, the coefficients of x2 and y2 cannot be reduced to 1.
Exercise 1.3
Find the general form of the equation of the following circles given their centre
and radius.
1. (0, – 4) and r = 4.
2. (– 7, – 5) and r = 8.
3. (– 2, 2) and r = 2.
4.
3, 1
2 4
and r = 1 .
2
Example 4
Find the centre and radius of the circle represented by (x – 3)2 + (y + 9)2 = 16.
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Write this equation in standard form.
(x – 3)2 + (y – (– 9))2 = 42, where h = 3, k = – 9 and r = 4.
The circle has centre (3, – 9) and radius 4 units.
?
Find
the
centre
and
the
radius
of
the
circle
(x + 5)2 + y2 = 25.
Standard form: ..........................................................
h = ........
k = ........
r = ........
C(........……)
Solution
The standard form of the equation is (x – (– 5))2 + (y – 0)2 = 52.
Hence, h = – 5, k = 0, r = 5 and C(– 5, 0).
Exercise 1.4
Complete the following table by finding the centre and the radius of the circles
defined by each given equation.
Equation
Centre
Radius
1. (x – 7)2 + (y + 8)2 = 25
2. x2 + (y – 9)2 = 144
3. (x + 14)2 + (y – 10)2 = 100
4. x2 + y2 = 9
16
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Did you know that...
... the symbols that appear in certain formulas were not
always those we know today?
Consider the square root symbol ( 3 ) used
in the formula for calculating distance. The
one we use today is of Germanic origin and is
perhaps a deformation of the letter “r”. It has
been used since the 16th century. Prior to
this time, the symbol shown on the right
stood for the word radix (Latin for “root”) and Fig. 1.12 Old symbol for
square root
was employed for the first time by Leonardo
da Pisa in 1220.
The symbol π, which is used to find the area
or circumference of a circle, came into use in
England around 1700. In 1859, Benjamin
Pierce, a Harvard professor, suggested that
the symbol shown on the right be used to
designate the letter “pi”.
Fig. 1.13 Old symbol for the
letter “pi”
In another formula that is more or less related to the circle, there is the cubic root
symbol ( 3 3 ) which allows us to find the
radius of a sphere when its volume is known
3
V . In 1525, German mathematician
r=
π
Christoff Rudolff created the following symbol by running three radicals together. The Fig. 1.14 Old symbol for cusymbol that is used today is of French origin
and dates back to the 17th century.
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Let’s now try to find the centre and radius of a circle using the general form of
its equation.
To do this, the general form of the equation must be rewritten in standard form
by completing the square. In fact, the standard equation contains two perfect
squares, that is, (x – h)2 and (y – k)2.
A perfect square is a polynomial with identical factors. An example of a perfect
square is x2 + 4x + 4 = (x + 2)(x + 2) = (x + 2)2.
The general form of a perfect square is:
x2 + 2ax + a2 = (x + a)2
x2 – 2ax + a2 = (x – a)2
Consider the form x2 + 2ax + a2. The coefficient of x is 2a and the constant term
is a2. The “key” to completing the square is to find the relationship between 2a
2
and a2. Divide 2a by 2 and square the result to obtain a2. Hence, 2a
= a2.
2
?
Complete the following expressions so as to obtain perfect squares.
x2 + 6x ..............
x2 – 3x ..............
y2 + 8y ..............
The resulting perfect squares are, respectively:
x 2 – 3x + 9
x2 + 6x + 9
4
since 6
2
2
= 32 = 9
–3
2
2
= 9
4
1.20
and
y2 + 8y + 16
8
2
2
= 42 = 16
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Example 5
Find the centre and the radius of the circle x2 + y2 – 8x + 4y + 11 = 0.
• Group the terms in x and the terms in y, and place the constant term on
the right-hand side of the equation.
(x2 – 8x + ....) + ( y2 + 4y + ....) = – 11
• Complete the square for the terms in x and for the terms in y and balance
the equation.
To balance an equation, add the same quantity to both sides
of the equation.
x 2 + 8x + – 8
2
2
+ y2 + 4y – 4
2
2
= – 11 + – 8
2
2
+ 4
2
2
(x2 – 8x + 42) + ( y2 + 4y + 22) = – 11 + 16 + 4
• Write the equation in the standard form.
(x – 4)2 + (y + 2)2 = 9 or (x – 4)2 + (y – (– 2))2 = 32
• Find the centre and the radius: C(4, – 2) and r = 3.
N.B. If, in the equation of the circle, the coefficient of x2 and of y2 is a number
other than 1, divide all the terms in the equation by this coefficient before
completing the square.
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To find the centre (h, k) and the radius r of a circle
given the general form of its equation:
1. group the terms in x and the terms in y, and place the
constant term on the right-hand side of the equation:
x2 + Dx + ... + y2 + Ey + ... = – F;
2. complete the square for the terms in x and for the terms in
y, and balance the equation:
x 2 + Dx + D
2
2
+ y 2 + Ey + E
2
2
=–F+ D
2
2
+ E
2
2
;
3. write the equation in standard form:
(x – h)2 + (y – k)2 = r2;
4. find the center C(h, k) and the radius r.
Apply this work method to the following example.
?
Find the centre and radius of the circle 4x2 + 4y2 + 8x – 4y – 11 = 0 by completing
the following steps.
First, divide all the terms by 4 to obtain an equation of the form
x2 + y2 + Dx + Ey + F = 0.
x2 + y2 + ...... x – ...... y – ...... = 0
Then, follow the steps outlined above.
1. (x2 ................) + (y2 ................) = ........
2.
?
x 2 + 2x + .....
2
?
x 2 + 2x + .....
?
+ y 2 – y + .....
?
+ y 2 – y + .....
2
? + ?
= 11 + .....
.....
4
? + ?
= 11 + .....
.....
4
3. (x + .....)2 + (y – .....)2 = ..... or (x – ........)2 + (y – .......)2 = .....2
4. C(...............) and r = .....
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Your solution is correct if it matches the solution shown below.
The equation of the form x2 + y2 + Dx + Ey + F = 0 is x 2 + y 2 + 2x – y – 11 = 0.
4
1. (x2 + 2x + ..…) + (y2 – y + ...…) = 11
4
2.
x 2 + 2x + 2
2
2
+ y2 – y + – 1
2
(x 2 + 2x + (1) 2) + y 2 – y + 1
2
2
3. (x + 1) + y – 1
2
4. C – 1, 1
2
2
2
2
= 11 + 2
4
2
2
+ –1
2
2
= 11 + 4 + 1
4
4 4
2
or (x – (– 1)) + y – 1
2
2
= 22
and r = 2
Now do the following exercises.
Exercise 1.5
Write the following equations in standard form and find the centre and radius
of each circle.
1. x2 + y2 – x + 4y + 4 = 0
C(..............) and r = ............
2. x2 + y2 + 10x = 0
C(..............) and r = ...........
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3. 4x2 + 4y2 + 20x – 24y – 3 = 0
C(..............) and r = ............
4. x2 + y2 + 3x = 0
C(..............) and r = ...........
Now it's time to put into practice what you have learned in this unit.
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PRACTICE EXERCISES
1. On each set of axes below, graph the circle obtained by applying the required
translation and give its equation.
y
Translation of (– 2, 4) units.
Equation:
..........................................................
1
1
x
2. Find the equation of the circle obtained by applying the required translation
and graph it.
y
Translation of (– 2, – 1) units applied to the circle x2 + y2 = 16.
Equation:
1
..........................................................
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3. What is the general form of the equation of the circle with centre
C(8, 0) and radius 5?
4. Find the general form of the equation with centre C(0, – 3) and radius 3 units.
5. Find the centre and the radius of the circles defined by the equations given
below.
a) x2 + y2 – 10x + 4y + 13 = 0
b) x2 + y2 – 2x + 2y – 7 = 0
6. For each of the following circles, find the standard form of the equation, the
centre and the radius.
a) x 2 + y 2 – x – y – 1 = 0
2
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b) 9x2 + 9y2 – 18x + 6y – 26 = 0
c) x2 + y2 + 14x + 24 = 0
d) x2 + y2 + 4x – 10y – 35 = 0
7. What is the general form of the equation of the circle with centre C 1 , – 3
2 4
1
and radius
of a unit?
4
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REVIEW ACTIVITY
1. What does r represent in the equation of the circle?
...........................................................................................................................
2. Complete the following sentences.
a) The equation of a circle with radius r and centred at the origin is ........ .
b) The equation (x – h)2 + (y – k)2 = r2 is that of the circle whose equation is
………………………… and to which a translation of .......……units has
been applied.
c) If the equation of a circle is (x – h)2 + (y – k)2 = r2, the centre of the circle
is ............. and its radius is ....... .
d) The general form of the equation of a circle is ............................................. .
e) To find the general form of the equation of a circle with centre (h, k) and
radius r:
1. write the equation of the circle in its standard form : .........................;
2. find the ............................;
3. simplify like terms and arrange them so as to obtain the equation of the
form ............................................... .
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f) To find the centre (h, k) and the radius r of a circle given the general form
of its equation:
1. group the terms in .......... and the terms in ........., and place the
............................ term on the right-hand side of the ...................... :
x2 + Dx + ... + y2 + Ey + ...… = – F;
2. complete the ............... for the terms in x and for the terms in y, and
................... the equation;
3. write the equation in the .......................... form: ................................. ;
4. find the centre ..................... and the radius ......... .
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THE MATH WHIZ PAGE
Let's Start with the Diameter
If A(2, 7) and B(– 4, 15) are the endpoints of the diameter of a circle,
find the general form of the equation of this circle.
N.B. To solve this problem, you need only find the circle's centre and
radius.
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