SCHOLAR Study Guide CfE Higher Mathematics Course Materials Topic 9: Circles Authored by: Margaret Ferguson Reviewed by: Jillian Hornby Previously authored by: Jane S Paterson Dorothy A Watson Heriot-Watt University Edinburgh EH14 4AS, United Kingdom.
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SCHOLAR Study Guide CfE Higher Mathematics … · CfE Higher Mathematics Course Materials Topic 9: Circles ... TOPIC 1. CIRCLES 3 ... The equation is now in the form x2 + y2 = r2
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SCHOLAR Study Guide
CfE Higher Mathematics
Course MaterialsTopic 9: Circles
Authored by:Margaret Ferguson
Reviewed by:Jillian Hornby
Previously authored by:Jane S Paterson
Dorothy A Watson
Heriot-Watt University
Edinburgh EH14 4AS, United Kingdom.
First published 2014 by Heriot-Watt University.
This edition published in 2016 by Heriot-Watt University SCHOLAR.
Members of the SCHOLAR Forum may reproduce this publication in whole or in part foreducational purposes within their establishment providing that no profit accrues at any stage,Any other use of the materials is governed by the general copyright statement that follows.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval systemor transmitted in any form or by any means, without written permission from the publisher.
Heriot-Watt University accepts no responsibility or liability whatsoever with regard to theinformation contained in this study guide.
Distributed by the SCHOLAR Forum.
SCHOLAR Study Guide Course Materials Topic 9: CfE Higher Mathematics
1. CfE Higher Mathematics Course Code: C747 76
AcknowledgementsThanks are due to the members of Heriot-Watt University's SCHOLAR team who planned andcreated these materials, and to the many colleagues who reviewed the content.
We would like to acknowledge the assistance of the education authorities, colleges, teachersand students who contributed to the SCHOLAR programme and who evaluated these materials.
Grateful acknowledgement is made for permission to use the following material in theSCHOLAR programme:
The Scottish Qualifications Authority for permission to use Past Papers assessments.
The Scottish Government for financial support.
The content of this Study Guide is aligned to the Scottish Qualifications Authority (SQA)curriculum.
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1
Topic 1
Circles
Contents
9.1 Looking back at the circle from National 5 . . . . . . . . . . . . . . . . . . . . . 3
We have formed two pairs of useful alternate angles. Notice in the following diagramthat the three angles of the triangle come together to form a straight line. The sum ofthe angles in any triangle equal 180◦. The angles which form a straight angle are calledsupplementary angles and equal 180◦.
A circle with its centre is drawn and a diameter drawn through the centre. A point ischosen on the circumference of the circle. . . not the end point of the existing diameter.A diameter can always be drawn from this point.
These two diameters are equal and bisect each other. . . the configuration of thediagonals is a rectangle. So given any diameter, any point on the circumference is acorner of a rectangle using the diameter as a diagonal.
The diameter subtends an angle of 90◦ at any point on the circumference. The rightangle is always on the circumference of the circle opposite the diameter and we call thisproperty "The angle in a semi-circle is a right angle".
A circle with its centre is drawn and a diameter drawn through the centre. A chord thatcuts the circle in two places is drawn at right angles to the diameter. The diameter is anaxis of symmetry. . . the chord is bisected.
A second chord parallel to the first is drawn to the right. . . the two points of intersectionare closer together and the chord is at right angles to the diameter. A third chord parallelto the first two is drawn further to the right. . . the two points of intersection are evencloser together and the chord is at right angles to the diameter.
At some point the two points of intersection will become one. . . the chord will becomea tangent and the tangent is at right angles to the diameter. A tangent will be at rightangles to a radius drawn to the point of contact.
• Look for isosceles triangles formed with two radii and a chord.
• The right angle is always on the circumference of the circle opposite thediameter and we call this property ’The angle in a semi-circle is a rightangle’.
• The diameter and tangent are perpendicular and we call this property ’Atangent is at right angles to the radius at the point of contact’.
Properties of a circle practice
Go online
Q1:
AB is a diameter and C is a pointon the circumference of the circle.
The study of the circle goes back beyond the recorded history. Euclid’s Elements BookIII, deals with the properties of circles.
A circle is the set of all points (or locus) in a plane that are equidistant from a centralpoint. In this topic we will look at different way of describing the equations of circles.
1.2.1 The equation of a circle with centre the origin
Circle radius
Go onlineThe easiest circle to construct is a circle with centre O, the origin. Here are someexamples.
(x − 1)2 + (y + 3)2 = 36 is the equation of the circle and the points P and Q lie onthe circumference.If the line through P, Q and the centre of the circle is parallel to the y-axis, find thecoordinates of P and Q.
Q23: A gearing system has two wheels, one large and one small.The line of centres of the two wheels is parallel to the x-axis and the equation of thelarger wheel is (x − 2)2 + (y + 4)2 = 64.Find the equation of the smaller wheel which lies to the right of the larger wheel and hasa radius of half of that of the larger wheel.
Q24: A child’s bicycle has two identical wheels with a clearance between them of 15cm.The bike measures 165 cm long when held against a wall (with the ground as the x-axis).Find the equations of the two circles which represent the wheels. Give your answer inthe form (x − a)2 + (y − b)2 = r2.
With a clearance of at least 1 cm the edge of the second wheel is at least 4 cm away;but the centres are 18 cm apart so the radius of the second wheel must be at the most14 cm.
The second wheel radius has a range of 0 < r ≤ 14.
b)If the clearance is then set at 11 cm, the second wheel has a radius of 4 cm.
The centre of the first wheel is (-2,-1).The x-coordinate of the centre of the second wheel will be −2 + 18 = 16.The y-coordinate is the same for both wheels at y = − 1.
The centre of the second wheel is (16,-1) with radius 4.The equation for the second wheel is (x − 16)2 + (y + 1)2 = 16.
• if b2 − 4ac = 0 then the line is a tangent to the circle;
• if b2 − 4ac > 0 then there are 2 points of intersection;
• if b2 − 4ac < 0 then the line does not meet the circle at all.
Key point
A radius and tangent are perpendicular at the point of contact.To find the gradient of the tangent we can therefore use the relationship m tangent ×mradius = −1.
First we must check that the point lies on the circumference of the circle by substitutingx = 7 and y = 9 in the equation of the circle72 + 92 − 4 × 7 − 10× 9 − 12 = 0 hence (7,9) lies on the circle.
2g = − 4 so g = − 2 and 2f = − 10 so f = − 5.
The centre of the circle is (−g,−f ) which in this case is (2,5).
To find the gradient of the tangent we must first find the gradient of the radius.
mradius = 9 − 57 − 2 = 4
5
The radius is perpendicular to the tangent so m tangent = −54 .
Beware: There are times where one circle could lie completely within another circle andso do not intersect.
Using the approach shown in the last example works in some cases only. It will not workwhen the sum of the distance between the centres and the radius of the smaller circleis less than the radius of the larger circle.
In the diagram below the radius of the larger circle is 5, the radius of the smaller circleis 1 and the distance between the centres is 3.
The distance between the two centres plus the radius of the smaller circle is 4 and thisis less than the radius of the larger circle. Therefore the small circle is completely insidethe large circle and they do not intersect.
Always make a sketch to help visualize the problem.
Examples
1. Problem:
Show that the circle P with equation (x − 4)2 + (y − 2)2 = 36 touches the circle Qwith equation (x − 7)2 + (y − 2)2 = 9 and find the point of contact.
Solution:
Circle P has centre (4,2) and radius 6.
Circle Q has centre (7,2) and radius 3.
The distance between the two centres is√
(7 − 4)2 + (2 − 2)2 =√9 = 3
Also notice that the centres of the circles both lie on the same horizontal line.
The centre of P is (4,2) so the circumference must lie 6 units from this point i.e. (10,2).The centre of Q is (7,2) and the circumference must lie 3 units to the right of it i.e. (10,2).Hence the point of contact of circles P and Q is (10,2) and the circles touch internally.
Find the coordinates of the points of intersection of circle V with equation (x − 2)2 +(y − 3)2 = 16 and circle W with equation (x + 1)2 + (y − 3)2 = 25.
Solution:
The points of intersection can be found by solving the two equations simultaneously.
Q38: Find the coordinates of the points of intersection of circle C with equation (x −3)2 + (y + 1)2 = 16 and circle D with equation (x − 3)2 + (y − 2)2 = 25.
• The equation of a circle with centre (0,0) and radius r is x2 + y2 = r2.
• The equation of a circle with centre (a, b) and radius r is (x − a)2 + (y − b)2 = r2.
• The general equation of the circle is x2 + y2 + 2gx + 2fy + c = 0, with centre(-g,-f) and radius
√g2 + f2 − c, provided g2 + f2 − c > 0.
Lines and circles
• The relationship between a line and a circle can be found by:
1. substituting the equation of the line into the equation of the circle;2. collecting like terms to obtain a quadratic equation;
3. evaluating and interpreting the discriminant:◦ if b2 − 4ac = 0 then the line is a tangent to the circle;◦ if b2 − 4ac > 0 then there are 2 points of intersection;◦ if b2 − 4ac < 0 then the line does not meet the circle at all.
• The coordinates of any points of intersection can be found by solving thequadratic.
• A radius and tangent are perpendicular at the point of contact.
• The equation of a tangent to a circle at the point (a, b) can be found by:
1. finding the gradient of the radius;
2. using mtangent ×mradius = − 1 to find the gradient of the tangent;
3. substituting the gradient of the tangent and the coordinates of the point ofcontact into y − b = m(x − a).
• It is useful to remember that the angle in a semi-circle is a right angle.
• It is useful to remember that a radius bisects a chord at right angles.
Intersecting circles
• Circles which do not touch can be one inside the other or completely apart.
• What are the coordinates of the centre of the larger circle? The centre of the largercircle is (2,-4).
• What are the coordinates of the centre of the smaller circle? The centre of thesmaller circle is at (14, -4).
Answer: The equation of the smaller wheel is (x − 14)2 + (y + 4)2 =16.
Q24:
Steps:
• What is the radius of the wheels? 37·5 cm
• What is the x-coordinate of the centre of the second wheel (wheel B)? 127·5
Answer:The first wheel A has centre at (37·5, 37·5) since it rests against the wall and the ground.It has equation (x − 37 · 5)2 + (y − 37 · 5)2 = 1406 · 25The second wheel B has centre at (127·5, 37·5). It has equation (x − 127·5)2 + (y − 37 · 5)2 = 1406 · 25
The general equation of a circle exercise (page 19)
Q25:
a) The centre is given by (2,1) and the radius is 3.
b) The centre is given by (-3,2) and the radius is 4.
c) The centre is given by (0,-4) and the radius is 5.
d) The centre is given by (5,0) and the radius is 2.
Q26:
a) The centre is given by (-9,0) and the radius is 14.
b) The centre is given by (0,6) and the radius is 5.
c) The centre is given by (-6,9) and the radius is 8.
d) The centre is given by (-8,3) and the radius is 8.
• To find the point(s) where the line and tangent intersect you must substitute for ythe equation of the line into the equation of the circle x2 + y2 − 4x − 16 = 0.
• The line and the tangent intersect when x = ? x = − 2
Answer: Yes, the line is a tangent to the circle.
Q48:
Steps:
• In order to find the gradient of the tangent, you must first find the gradient of radiusCP.
• What is the gradient of radius CP? − 45
• What is the gradient of the tangent at P? 54
Answer: y = 54x + 1
Q49:
Steps:
• The circle has centre (a, b).What are a and b? (-3,3)
• What is the gradient of the radius from the centre to the point (-7,0)? 34
• What is the gradient of tangent T? − 43
Answer: y = − 43x − 28
3
Q50: (3,3)
Q51:
Steps:
• What is the distance between the centres of circles A and B? 2√13
• Compare the distance between the centres with the radii.
Answer: Circle A does not touch or intersect circle B because the sum of the radii isless than the distance between the centres of the circles.
Q52:
Steps:
• What is the distance between the centres of circles A and B?√2
• What is the distance between the centres + radius of the smaller circle? 3·414
• What is the radius of the larger circle? 4
• Consider whether circle Q could lie completely inside circle P.
Answer: Circle P and circle Q do not touch or intersect because the distance betweenthe two centres + the radius of the smaller circle is less than the radius of the largercircle.