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 1 Preparation for the Mathematics examination brought to you by Kagiso Trust 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. 2 Analytical geometry of the circle REVISION UNIT 14: ANALYTICAL GEOMETRY OF THE CIRCLE The circle is defined as the locus of all points, P ( x; y ) , which are equidistant from some given point C
(a; b) suppose that the distance of the points P from the given point C is r , then CP 2 = r 2 and using the distance formula: ( x − a )2 + ( y − b )2
= r2 which gives the STANDARD FORM of the equation of a circle with centre (a; b ) and radius r . In CANONICAL FORM the equation of the circle is x 2 + y 2 + 2 gx + 2 fy + c = 0 . 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)
( x + 2 )2 + ( y − 3 ) 2
b)
x 2 + y 2 − 10 x − 6 y = −18 = 49 2. Show that part of the line 4 y − 3 x − 9 = 0 is a chord of the circle x 2 + y 2 − 2 x − 6 y − 15 = 0
and find the length of this chord. 3. Find the equation of (i) the tangent and (ii) the normal to the circle x 2 + y 2 + 4 x − 10 y = −4 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 x 2 + y 2 − 2 y − 4 = 0 and x 2 + y 2 − x + y − 12 = 0 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 x 2 + y 2 + 2 x − 8 y + 5 = 0 and x 2 + y 2 − 4 x − 4 y + 7 = 0 are orthogonal (meaning that their radii meet at right angles at the points where the circles intersect). Preparation for the Mathematics examination brought to you by Kagiso Trust ANSWERS
Answers: Equation of the circle is x 2 + y 2 − 2 x − 6 y − 15 = 0 or 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 4 y + 3 x = 38 Equation of the normal is 3 y − 4 x = 32 4. HINT: You can use either form of the equation of a circle. But it will be quick to use the canonical form: x 2 + y 2 + 2 fx + 2 gy + c = 0 . Substitute each of the given points into this equation so you will have 3 simultaneous equations in 3 unknowns: f , g and c . Taking these equations two at a time, eliminate c
by subtraction. That will leave you with two simultaneous equations in the two unknowns f and g . Solve for f and g . ( x − 1) + ( y − 3) = 25 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: 3 y − x − 8 = 0 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? 2
4 2
Analytical geometry of the circle PAPER 2 QUESTION 2 DoE/ADDITIONAL EXEMPLAR 2008 Preparation for the Mathematics examination brought to you by Kagiso Trust PAPER 2 QUESTION 2 Number Hints and answers 2.1 Work out r and write down the equation. Answer: x 2 + y 2 = 25 2.2 AB is what to the circle? So its length? Answer: AB = 10 units 2.3 Multiply out the brackets. Answer: x 2 − 6 x + y 2 + 8 y − 75 = 0 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 180 o about the origin 2.5 Use the gradient formula. Answer: m AB = −
2.6 Answer: y =
2.7 29
3
6 3
25
x−
4
4
You know something about C. Substitute it into the equation of BC you found in 2.6 above. Answer: k =
Work out the solutions in the boxes below 4
3
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. DoE/ADDITIONAL EXEMPLAR 2008 Analytical geometry of the circle PAPER 2 QUESTION 2 DoE/NOVEMBER 2008 Preparation for the Mathematics examination brought to you by Kagiso Trust PAPER 2 QUESTION 2 Number Hints and answers 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: x 2 + y 2 = 169 2.2 You know the y ‐intercept and you can work out the gradient of the line. Answer: y =
2.3 2.4 Work out the solutions in the boxes below 5
x 12
Simple! The point P is symmetrical with the point Q; or P is a rotation of Q through 180 o .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. You know the measure of mOQ from 2.2. You are given that QR is a tangent you also should know that diameter ⊥ tangent. So use m OQ .m QR = −1 . 2.5 2.6 2.7 Answer: You write down the answer. You know mQR from 2.4 and you are given a point on QR. So you can write the equation in the form y = mx + c
and work out c. Answer: y = −2,4 x + 33,8 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 Simple! This is a translation of the centre of the circle in the diagram. Answer: You write down the answer. 8 DoE/NOVEMBER 2008 Analytical geometry of the circle MORE QUESTIONS FROM PAST EXAMINATION PAPERS
Exemplar 2008
Preparation for the Mathematics examination brought to you by Kagiso Trust Preparatory Examination 2008
10 Analytical geometry of the circle Feb – March 2009
Preparation for the Mathematics examination brought to you by Kagiso Trust November 2009 (Unused paper)
12 Analytical geometry of the circle November 2009 (1)
Preparation for the Mathematics examination brought to you by Kagiso Trust Feb – March 2010
14 Analytical geometry of the circle 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; 5
) 2
2
5⎞
125
⎛
2.4 (x − 3) + ⎜ y − ⎟ =
2⎠
4
⎝
2.5 y = −2 x + 21 2
2.6 A’(‐2; 5) Preparatory Examination 2008 2.1 k = 3 2.2.1 M(3; ‐1) 2.2.2 r = 10 − t 2.2.3 t = −7 Feb/March 2009 2.1 Midpoint of AB = (‐4; 3) 2.2 y = −2 x + 5 2.3 AM = 20
( ) 2.4 x 2 + y 2 + 8 x − 6 y + 5 = 0 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 ( x − 3) + ( y + 2 ) = 25 2
2
2.1.4 θ = 36,87 o 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. 4
31
x+ 3
3
2
2
2.2.1 (x + 3) + ( y − 6 ) = 10 2.1.7 y = −
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 θ = 45 o 5.7 r = 2,93 Feb/March 2010 6.1 Centre at (‐4; ‐2) r = 58 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.