Analytical Geometry of the Circle

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MATHEMATICS

Learner’s Study and

Revision Guide for

Grade 12

CIRCLE GEOMETRY

Revision Notes, Exercises and Solution Hints by

Roseinnes Phahle

Examination Questions by the Department of Basic Education

Preparation for the Mathematics examination brought to you by Kagiso Trust

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Contents

Unit 14

All you need to know: Revision notes 3

Exercise 14 3

Answers 4

Examination questions with solution hints and answers 5

More questions from past examination papers 9

Answers 15

How to use this revision and study guide

1. Study the revision notes given at the beginning. The notes are interactive in that in some parts you are required to make a response based on your prior learning of the topic from your teacher in class or from a textbook. Furthermore, the notes cover all the Mathematics from Grade 10 to Grade 12.

2. “Warm‐up” exercises follow the notes. Some exercises carry solution HINTS in the answer section. Do not read the answer or hints until you have tried to work out a question and are having difficulty.

3. The notes and exercises are followed by questions from past examination papers.

4. The examination questions are followed by blank spaces or boxes inside a table. Do the working out of the question inside these spaces or boxes.

5. Alongside the blank boxes are HINTS in case you have difficulty solving a part of the question. Do not read the hints until you have tried to work out the question and are having difficulty.

6. What follows next are more questions taken from past examination papers.

7. Answers to the extra past examination questions appear at the end. Some answers carry HINTS and notes to enrich your knowledge.

8. Finally, don’t be a loner. Work through this guide in a team with your classmates.

Analytical geometry of the circle

REVISION UNIT 14: ANALYTICAL GEOMETRY OF THE CIRCLE

The circle is defined as the locus of all points, P ( )yx; , which are equidistant from some given point C

( )ba; suppose that the distance of the points P from the given point C is r , then

22CP r=

and using the distance formula:

( ) ( ) 222 rbyax =−+−

which gives the STANDARD FORM of the equation of a circle with centre ( )ba; and radius r .

In CANONICAL FORM the equation of the circle is 02222 =++++ cfygxyx .

To convert from canonical to standard form, use the method of completing the square and complete the squares on the x and y terms.

IN ORDER TO ANSWER THE EXAM QUESTION ON THE CIRCLE, YOU MUST KNOW THE ABOVE AS WELL AS ALL THE ANALYTICAL GEOMETRY OF THE STRAIGHT LINE WHICH INCLUDES THE DISTANCE

FORMULA, THE MID‐POINT FORMULA, THE PRODUCT OF THE SLOPES OF PERPENDICULAR LINES AND THE EQUATION OF A STRAIGHT LINE . LEARN ALL THIS AND YOU WILL BE ASSURED OF ANSWERING

TWO QUESTIONS IN PAPER 2!!!!

EXERCISE 14

1. Find the centre and radius of the circles with equations:

a) ( ) ( ) 4932 22 =−++ yx

b) 1861022 −=−−+ yxyx

2. Show that part of the line 0934 =−− xy is a chord of the circle 0156222 =−−−+ yxyxand find the length of this chord.

3. Find the equation of (i) the tangent and (ii) the normal to the circle 410422 −=−++ yxyx at

the point (2; 8) on the circumference of the circle. 4. Find the equation of the circle passing through the points (‐3; 6), (1;‐2) and (4; 7); find its radius

and centre as well.

5. Show that the circles 04222 =−−+ yyx and 01222 =−+−+ yxyx intersect in two

distinct points and find the equation of the common chord. (Hint: the common chord is found by subtracting one equation from the other. Thereafter, find the points of intersection.)

6. Prove that the circles 058222 =+−++ yxyx and 074422 =+−−+ yxyx are orthogonal

(meaning that their radii meet at right angles at the points where the circles intersect).


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ANSWERS

EXERCISE 14 1 (a) radius = 7 and centre at (‐2;3) (b) radius = 4 and centre at (5;3) 2. HINT: If the line is a chord of the circle then it must intersect the circle at two distinct points. To find out, solve the two equations simultaneously. Answer: The line cuts the circle at the points (‐3; 0) and (5; 6). Hence the line is a chord. 3. A normal refers to a perpendicular line. In this case, the perpendicular to the tangent passes through the origin of the circle and so the radius is a segment of the normal. Find the centre of the given circle and together with the point (2; 8) find the slope of the radius (which is the slope of the normal). HINT: Knowing the gradient of the normal and a point on the normal, you can determine the equation of the normal. You can deduce the gradient of the tangent from the gradient of the normal and so go on to determine the equation of the tangent. Answers: Equation of the tangent is 3834 =+ xy Equation of the normal is 3243 =− xy 4. HINT: You can use either form of the equation of a circle. But it will be quick to use the canonical

form: 02222 =++++ cgyfxyx . Substitute each of the given points into this equation so you will have 3 simultaneous equations in 3 unknowns: cgf and , . Taking these equations two at a time, eliminate cby subtraction. That will leave you with two simultaneous equations in the two unknowns

gf and . Solve for gf and .

Answers: Equation of the circle is

0156222 =−−−+ yxyx or

( ) ( ) 2531 22 =−+− yx The centre is at (1; 3) and the radius is 5. 5. On subtracting the equation of one circle from the equation of the other the equation of the common chord is found as: 083 =−− xy To find the points of intersection of the two circles, solve the equation of the chord simultaneously with any of the two equations of the circle. Other Answers: Points of intersection are (‐2; 2) and (1; 3). 6. HINT: Do as above to find the equation of the common chord. Use the equation of the common chord to find the points of intersection of the two circles. The points of intersection are (1,2; 2,0) and (2,3; 3,0) rounded off to 1 decimal place. Find the coordinates of the centre of each circle. You should get (2; 2) and (‐1; 4). Find the gradients of the radii drawn to the points of intersection. What is the product of the gradients? Are you now able to deduce perpendicularity?


PAPER 2 QUESTION 2 DoE/ADDITIONAL EXEMPLAR 2008


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PAPER 2 QUESTION 2 DoE/ADDITIONAL EXEMPLAR 2008

Number Hints and answers Work out the solutions in the boxes below 2.1 Work out r and write down the equation.

Answer: 2522 =+ yx

2.2 AB is what to the circle? So its length? Answer: AB = 10 units

2.3 Multiply out the brackets.

Answer: 07586 22 =−++− yyxx

2.4 Is this a transformation? If so describe it fully. Answer: these coordinates of A represent a rotation of B through an angle of o180 about the origin

2.5 Use the gradient formula.

Answer: 34

−=ABm

2.6 There are two formulae on the formula sheet. Choose either one. Recall that the product of the slopes of lines that are perpendicular is ‐1.

Answer: 425

43

−= xy

2.7 You know something about C. Substitute it into the equation of BC you found in 2.6 above.

Answer: 3

29=k


PAPER 2 QUESTION 2 DoE/NOVEMBER 2008


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PAPER 2 QUESTION 2 DoE/NOVEMBER 2008

Number Hints and answers Work out the solutions in the boxes below 2.1 You know the centre and you are

given a point on the circle. So work out the radius and write down the equation of the circle.

Answer: 16922 =+ yx

2.2 You know the y ‐intercept and you can work out the gradient of the line.

Answer: xy125

=

2.3 Simple! The point P is symmetrical with the point Q; or P is a rotation of Q through o180 .This alone should enable you to write the coordinates of P without any working out to do. Or, you can do so working out by solving the answers to 2.1 and 2.2 as simultaneous equations. Answer: You write down the answer.

2.4 You know the measure of OQm from

2.2. You are given that QR is a tangent you also should know that diameter ⊥ tangent. So use 1. −=QROQ mm .

Answer: You write down the answer.

2.5 You know QRm from 2.4 and you are

given a point on QR. So you can write the equation in the form cmxy +=and work out .c Answer: 8,334,2 +−= xy

2.6 There are several ways of calculating the value of t . One way is to substitute R( t ;1) into the equation of QR. Answer: t =14,5

2.7 Simple! This is a translation of the centre of the circle in the diagram. Answer: You write down the answer.


MORE QUESTIONS FROM PAST EXAMINATION PAPERS

Exemplar 2008


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Preparatory Examination 2008


Feb – March 2009


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November 2009 (Unused paper)


November 2009 (1)


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Feb – March 2010


ANSWERS

Exemplar 2008 2.1 Proof required. Provide a proof and check with the classmates or teacher if it is correct. 2.2 D(‐2; 0)

2.3 C(3; 25)

2.4 ( )4

125253

22 =⎟

⎠⎞

⎜⎝⎛ −+− yx

2.5 212 +−= xy 2.6 A’(‐2; 5) Preparatory Examination 2008 2.1 3=k 2.2.1 M(3; ‐1)

2.2.2 tr −= 10 2.2.3 7−=t Feb/March 2009 2.1 Midpoint of AB = (‐4; 3) 2.2 52 +−= xy

2.3 AM = 20 ( ) 2.4 056822 =+−++ yxyx 2.5 Say which one is it and prove it. November 2009 (Unused paper) 2.1.1 Midpoint of BD = (3; ‐2) 2.1.2 Proof required. Provide a proof and check with the classmates or teacher if it is correct.

2.1.3 ( ) ( ) 2523 22 =++− yx

2.1.4 o87,36=θ 2.1.5 E(6; ‐6) 2.1.6 Give an explanation and check it with your classmates or teacher to see if it is acceptable.

2.1.7 331

34

+−= xy

2.2.1 ( ) ( ) 1063 22 =−++ yx

2.2.2 Give an explanation and check it with your classmates or teacher to see if it is acceptable. November 2009(1) 5.1 M(‐1; 1) 5.2 Proof required. Provide a proof and check with the classmates or teacher if it is correct. 5.3 Give an explanation and check it with your classmates or teacher to see if it is acceptable. 5.4 AD = AB = 10 5.5 Proof required. Provide a proof and check with the classmates or teacher if it is correct. 5.6 o45=θ 5.7 93,2=r Feb/March 2010 6.1 Centre at (‐4; ‐2)

58=r 6.2 Distance = 11,31 6.3 Proof required. Provide a proof and check with the classmates or teacher if it is correct. 6.4 Proof required. Provide a proof and check with the classmates or teacher if it is correct.

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Analytical Geometry of the Circle