MEI Core 1 Coordinate Geometry Section 2: Curves and circles Study Plan Background This section focuses on circles, look...
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MEI Core 1 Coordinate Geometry Section 2: Curves and circles Study Plan Background This section focuses on circles, looking at their equations and properties.
Detailed work plan 1. Read pages 60 – 61, which look at some common curves. You need to be able to recognise curves of this form. 2. As an introduction to circles, look at the Circles dynamic spreadsheet (you will find some instructions in the Notes and Examples). Also look at the Flash resources Equation of a circle centre O and Equation of a circle centre (a, b). You may also find the Mathcentre video Coordinate geometry of a circle useful. 3. Read pages 61 – 66. It is important that you can find the centre and radius of a circle easily from its equation. Practice will help. There are some additional examples in the Notes and Examples. 4. It is also important to learn the facts about a circle on pages 63 and 64 as these will help you find the quickest way to a solution for some of the questions. These properties are demonstrated in the Flash resources Angle in a semicircle, Perpendicular to a chord, and Tangent and radius. 5. Exercise 2E Attempt questions 1, 2, 5*, 7*, 8, 10*, 12, 13* 6. For extra practice try the interactive questions Finding the radius and centre of a circle (circle equation in its simplest form), Finding the radius and centre of a circle (circle equation in its expanded form), Find the equation of a circle (from its centre and one point on its circumference) and Tangents and normals to circles. 7. Circles Activities includes two activities. The first involves matching up equations of circles with the appropriate diagram, and the second looks at finding the equation of the tangent to a circle. 8. Read pages 68 – 72. Solving simultaneous equations is revisited here but with equations of curves as well as lines. Some slightly different techniques are required so you should read the examples on pages 69 to 71 to make sure that you understand these. There are more examples in the Notes and Examples.
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MEI C1 Coordinate geometry Section 2 Study plan 9. You can also look at the Circles dynamic spreadsheet (Circle and a line) and the Flash resources Intersection of a curve and a line and Intersection of a circle and a line. 10. Exercise 2F Attempt questions 2, 3*, 5*, 7, 8*, 10, 11 For extra practice try the interactive questions Quadratic and line intersection and Circle and line intersection.
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MEI Core 1 Coordinate Geometry Section 2: Curves and circles Notes and Examples These notes and examples contain subsections on The equation of a circle Finding the equation of a circle Circle geometry The intersection of a line and a curve The intersection of two curves
The equation of a circle Start this section by looking at the Circles dynamic spreadsheet. Select the Circle Equations sheet. First, set the centre of the circle to be the origin and vary the radius. Look at how the equation of the circle changes. Now vary the coordinates of the centre of the circle, and look at how the equation of the circle changes. You can also explore equations of circles using the Flash resources Equation of a circle centre O and Equation of a circle centre (a, b).
You should find out the following results, which you need to learn: The general equation of a circle, centre the origin and radius r is x2 y 2 r 2 The general equation of a circle, centre (a, b) and radius r is ( x a)2 ( y b)2 r 2
Make sure you understand why these equations describe circles. See page 62 in the textbook for help.
Example 1 For each of the following circles find (i) the coordinates of the centre and (ii) the radius. (a) x² + y² = 49 (b) (x + 2)² + (y 6)² = 9
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MEI C1 Coordinate geom. Section 2 Notes & Examples Solution (a) x² + y² = 49 can be written as x² + y² = 7². (i) The coordinates of the centre are (0, 0) (ii) The radius is 7.
This is a particular case of the general form x² + y² = r² which has centre (0, 0) and radius r.
(b) (x + 2)² + (y 6)² = 9 can be written as (x (2))² + (y 6)² = 3². (i) The coordinates of the centre are (2, 6) (ii) The radius is 3. This is a particular case of the general form (x a)² + (y b)² = r² which has centre (a, b) and radius r.
For practice in examples like the one above, try the interactive questions Finding the radius and centre of a circle (circle equation in its simplest form).
The first activity in Circle Activities gives you practice in matching equations of circles with diagrams.
Sometimes the circle equation needs to be rearranged into its standard form before you can find the centre and radius.
Example 2 Show that the equation x2 y 2 4 x 6 y 3 0 represents a circle, and find its centre and radius. Solution The general equation of a circle is Multiplying out:
( x a)2 ( y b)2 r 2 x 2 2ax a 2 y 2 2by b 2 r 2
x 2 y 2 2ax 2by a 2 b 2 r 2 0 2a 4 a 2 Comparing with the original equation: 2b 6 b 3 a 2 b 2 r 2 3 4 9 r 2 3 r 2 16
( x 2)2 ( y 3)2 42 The equation can be written as This is the equation of a circle, centre (-2, 3), radius 4.
For practice in examples like the one above, try the interactive questions Finding the radius and centre of a circle (circle equation in its expanded form).
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MEI C1 Coordinate geom. Section 2 Notes & Examples In the example above, you are using the technique of completing the square, which is covered briefly in Chapter 1 (pages 19 – 20), and in more depth in Chapter 3 (pages 98 – 99). You may find the Mathcentre video Coordinate geometry of a circle useful.
Finding the equation of a circle In Section 1 you looked at different ways of finding the equation of a line. You can find the equation of a line from the gradient and the intercept, or from the gradient and one point on the line, or from two points on the line. In the same way, there are several ways of finding the equation of a circle, depending on the information available.
Finding the equation of a circle from the radius and centre
Example 3 Find the equation of each of the following. (a) a circle, centre (0, 0) and radius 4. (b) a circle, centre (3, 4) and radius 6. Solution (a) The equation of a circle centre the origin is x² + y² = r² r = 4 so the equation is i.e.
x² + y² = 4² x² + y² = 16
(b) The equation of a circle centre (a, b) and radius r is (x – a)² + (y – b)² = r² a = 3, b = 4 and r = 6 so the equation is i.e.
(x 3)² + (y (4))² = 6² ( x 3)² ( y 4)² 36
Finding the equation of a circle from its centre and one point on its circumference If you know the centre of the circle and one point on its circumference, you can find the radius by calculating the distance between these two points. You can then find the equation of the circle.
Example 4 Find the equation of the circle, centre (1, -2), which passes through the point (-2, -3). Solution The distance r between (1, -2) and (-2, -3) is given by:
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MEI C1 Coordinate geom. Section 2 Notes & Examples r
1 (2) 2 (3) 2
2
32 12 10 The radius of the circle is therefore 10 . The equation of the circle is ( x 1)2 ( y 2)2 10
For practice in examples like the one above, try the interactive questions Find the equation of a circle.
Finding the equation of a circle from three points on its circumference To find the equation of a line, you need the coordinates of two points on the line. To find the equation of a circle, you need the coordinates of three points on the circumference of the circle. One method is illustrated by the Circles dynamic spreadsheet. Select the sheet Circumcentre and follow the instructions on the sheet. This demonstration shows that the centre of the circle is the intersection of the perpendicular bisector of each pair of points. To find the centre of a circle through three points A, B and C, it is sufficient to find two of the perpendicular bisectors. For example, you can find the equations of the perpendicular bisectors of AB and BC, and then solve these equations simultaneously to find the point of intersection, i.e. the centre of the circle. You can then use the coordinates of the centre and one of the three points A, B and C to find the radius of the circle (as in Example 4), and hence find the equation of the circle.
Example 5 Find the equation of the circle passing through A (1, 3), B (9, 1) and C (8, 4). Solution C (8, 4)
B (9, 1)
A sketch is often helpful. The sketch does not need to be accurate. It gives some idea of roughly where the centre is, so you can check your answer is reasonable.
A (1, -3)
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MEI C1 Coordinate geom. Section 2 Notes & Examples You want to find the equation of the perpendicular bisector of AB. This is perpendicular to AB and passes through the midpoint M of AB.
y 2 y1 x 2 x1 1 ( 3) 4 1 = = m 91 8 2
The gradient of AB is found by using m
Note: Looking at the sketch we expect the gradient of AB to be positive.
Using m1m2 1 , the gradient of the perpendicular bisector is 2.
x x y y The midpoint M of AB is found by using M = 1 2 , 1 2 . 2 2 1 9 3 1 You are given A(1, -3) and B(9, 1) so M is , = (5, 1) 2 2 The perpendicular bisector is found using y y1 m( x x1 ) with ( x1 , y1 ) = (5, 1) and m = 2. so y ( 1) 2( x 5) y + 1 = 2x + 10 y = 2x + 9 (equation I) Next, use the same method to find the perpendicular bisector of BC.
The gradient of BC is
4 1 3 89
Note: Looking at the sketch, we expect the gradient of BC to be negative.
Therefore the gradient of the perpendicular bisector of BC is
1 . 3
9 8 1 4 , The midpoint N of BC is so N is (8.5, 2.5). 2 2 The equation of the perpendicular bisector is y 2.5 = 13 (x 8.5) 3(y 2.5) = x 8.5 3y 7.5 = x 8.5 3y = x – 1 (equation II)
y = 2x + 9 (equation I) 3y = x – 1 (equation II) Substituting (I) into (II)
Next, find the coordinates of the centre of the circle by solving equations (I) and (II) simultaneously.
3(2x + 9) = x – 1
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MEI C1 Coordinate geom. Section 2 Notes & Examples 6x + 27 = x – 1 28 = 7x x=4 Substituting x = 4 into equation (I) gives y = 2(4) + 9 = 1 So the coordinates of the centre are (4, 1).
Note: Looking at the sketch this appears to be a plausible result.
The radius is the distance between the centre (4, 1) and a point on the circumference such as (9, 1). This can be found by using d ( x1 x 2 ) 2 ( y1 y 2 ) 2 . radius = ( 9 4 ) 2 (1 1) 2 =
25 = 5
Finally, using the general form (x a)² + (y b)² = r² with a = 4, b = 1 and r = 5 the equation of the circle is (x 4)² + (y 1)² = 25. Note: You should check that each of the points A, B and C satisfy this equation.
Circle geometry The three facts about circles given on pages 63 and 64 are important. They often help to solve problems involving circles. 1.
The angle in a semicircle is a right angle. See the Flash resource The angle in a semicircle for a demonstration.
2.
The perpendicular from the centre of a circle to a chord bisects the chord. See the Flash resource Perpendicular to a chord for a demonstration.
3.
The tangent to a circle is perpendicular to the radius at that point See the Flash resource Tangent and radius for a demonstration.
Keep these properties in mind when dealing with problems involving circles. For some practice in using the third property, try the interactive questions Tangents and normals to circles.
The second activity in Circles activities looks at using the third property to find the equation of the tangent to a circle.
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MEI C1 Coordinate geom. Section 2 Notes & Examples The intersection of a line and a curve Just as the point of intersection of two straight lines can be found by solving the equations of the two lines simultaneously, the point(s) of intersection of a line and a curve can be found by solving their equations simultaneously. In many cases, the equations of both the line and the curve are given as an expression for y in terms of x. When this is the case, a sensible first step is to equate the expressions for y, as this leads to an equation in x only.
Example 6 Find the coordinates of the points where the line y = x + 2 meets the curve y = x² 3x + 5. Solution x² 3x + 5 = x + 2 x² 4x + 3 = 0 (x 3)(x 1) = 0 x = 3 or x = 1 When x = 3 then y = 3 + 2 = 5 When x = 1 then y = 1 + 2 = 3.
Equate the expressions for y to give an equation in x only
Substitute the x values into the equation of the line
The points where the line meets the curve are (3, 5) and (1, 3).
You should check that each of these points satisfies the equation of the curve. (You have already used the equation of the line to find the y-values).
Notice that this problem involved solving a quadratic equation, which in this case had two solutions, showing that the line crossed the curve twice. However, the quadratic equation could have had no solutions, which would indicate that the line did not meet the curve at all, or one repeated solution, which would indicate that the line touches the curve. You can look at some examples with the Flash resource Intersection of a curve and a line. For practice in examples like the one above, try the interactive questions Quadratic and line intersection.
The next example looks at the intersection of a line and a circle. Before reading this example, look at the Circles dynamic spreadsheet. Select the sheet Circle and a line. Try varying the equation of the line and/or the circle, and make sure that you can see that there may be two intersections, no intersections or one intersection (in which case the line touches the circle).
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MEI C1 Coordinate geom. Section 2 Notes & Examples You can also look at the Flash resource Intersection of a circle and a line.
Example 7 Find the coordinates of the point(s) where the circle ( x 2)2 ( y 1)2 9 meets (i) the line y = 5 (ii) the line x = 1 (iii) the line y = 2 – x Solution (i) Substituting y = 5 into the equation of the circle: ( x 2) 2 (5 1) 2 9
( x 2) 16 9 2
The expression (x + cannot be negative
2)²
( x 2) 2 7 There are no solutions. The line does not meet the circle. (ii)
Substituting x = 1 into the equation of the circle: (1 2) 2 ( y 1) 2 9
9 ( y 1) 2 9 ( y 1) 2 0 y 1 The line touches the circle at (1, 1). (iii)
The point is on the line x = 1, so its x-coordinate must be 1.
Substituting y = 2 – x into the equation of the circle: ( x 2) 2 (2 x 1) 2 9
( x 2) 2 (1 x) 2 9 x2 4x 4 1 2x x2 9 2x2 2 x 4 0 x2 x 2 0 ( x 1)( x 2) 0 x 1 or x 2
Substitute the x values into the equation of the line to find the y-coordinates.
When x = 1, y = 2 – 1 = 1 When x = –2, y = 2 – (–2) = 4 The line crosses the circle at (1, 1) and (–2, 4).
For practice in examples like the one above, try the interactive questions Circle and line intersection.
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MEI C1 Coordinate geom. Section 2 Notes & Examples The intersection of two curves As before, the intersections of two curves can be found by solving the equations of the curves simultaneously. As in Example 4, in many cases a sensible first step is to equate the expressions for y.
Example 8 Find the coordinates of the points where the curve y = x² 6x + 5 intersects the curve y = 2x² + 12x 19. Equate the expressions for y to give an equation in x only
Solution x² 6x + 5 = 2x² + 12x 19 3x² 18x + 24 = 0 x² 6x + 8 = 0 (x 2)(x 4) = 0 x = 2 or x = 4
There is a common factor of 3, so divide by 3 before factorising.
When x = 2, y = (2)² - 6(2) + 5 = 3 When x = 4, y = (4)² 6(4) + 5 = 3
Substitute the x values into the equation of one of the curves (in this case the first one).
The points of intersection are (2, 3) and (4, 3).
You should check that each of these points satisfy the equation of the second curve. (You have already used the equation of the first curve to find the y-values).
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MEI Core 1 Coordinate Geometry Section 2: Curves and circles Crucial points 1. Draw a diagram In most questions involving coordinate geometry, it is helpful to draw a sketch diagram. It does not need to be accurate, but it will help to give you a rough idea of the answer you might expect. 2. Make sure that you can recognise standard curves Look at the diagrams on pages 60 and 61 and make sure that you know the forms of these curves. 3. Make sure you know the standard circle equations The general equation of a circle, centre (0, 0) and radius r is: x2 y 2 r 2 The general equation of a circle, centre (a, b) and radius r is:
x a y b 2
2
r2
4. Finding the intersection of a line and a curve To find the coordinates of the point(s) where a line meets a curve, you solve the equations simultaneously. The condition for the line to be a tangent to the curve is that there is a repeated root. (For a line and a quadratic curve this means that the discriminant of the resulting quadratic equation is 0, i.e. b² - 4ac = 0). To find the coordinates of the point(s) where two curves meet you solve their equations simultaneously. 5. You are expected to know these circle properties: (i) the angle in a semicircle is a right angle (ii) the perpendicular from the centre of a circle to a chord bisects the chord (iii) the tangent to a circle at a point is perpendicular to the radius through that point These circle properties are often useful in examination questions. Keep them in mind when answering questions involving circles.
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MEI Core 1 Coordinate Geometry Section 2: Curves and circles Exercise Do not use a calculator in this exercise. 1. Find, in the form x 2 + y 2 + px + qy = c , the equation for each of the following circles. (i) centre (0, 0), radius 6 (ii) centre (3, 1), radius 5 (iii) centre (-2, 5), radius 1 (iv) centre (0, -4), radius 3 2. For each of these circles, write down the coordinates of the centre and the radius. (i) x 2 + y 2 = 100 (ii)
(iii) (iv)
( x − 2 ) + ( y − 7 ) = 16 2 2 ( x + 3) + ( y − 4 ) = 4 2 2 ( x + 4 ) + ( y + 5) = 20 2
2
3. For each of these circles, find the coordinates of the centre and the radius. (i) x 2 + y 2 + 4 x − 5 = 0 (ii)
x 2 + y 2 − 6 x + 10 y + 20 = 0
(iii) x 2 + y 2 − 2 x − 3 y + 3 = 0 4. The point C is (4, -2) and the point A is (6, 3). Find the equation of the circle centre C and radius CA. 5. The points A (2, 0) and B (6, 4) form the diameter of a circle. Find the equation of the circle. 6. A circle passes through the points Q(0, 3) and R(0, 9) and touches the x-axis. Work out two possible equations. 7. (i) Show that the line y = 4 − x is a tangent to the circle x 2 + y 2 = 8 . (ii) Show that the line 4 y = 3x − 25 is a tangent to the circle x 2 + y 2 = 25 . 8. Find the coordinates of the points where the following lines and parabolas intersect. (i) y = 3x +1 and y = x 2 − 4 x + 7 (ii) y = x − 2 and y = x 2 + 2 x − 8
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MEI C1 Coord. geom. Section 2 Exercise 9. The line 2 y + x = 10 meets the circle x 2 + y 2 = 65 at P and Q. Calculate the length of PQ. 10. The points P (-2, 6), Q (6, 0) and R (5, 7) all lie on a circle. (i) Show that PR is perpendicular to QR. (ii) Explain why the result from (i) shows that PQ is a diameter of the circle. (iii) Hence calculate the equation of the circle. 11. (i) (ii) (iii) (iv)
Write down the equation of the circle centre (0, 0) and radius 17 . Show that the point P(-4, -1) lies on the circle. Find the equation of the tangent at P. The line x + y = 3 meets the circle at two points, Q and R. Find the coordinates of Q and R. (v) Find the coordinates of the point, S, where the tangent at P intersects the line x + y = 3.
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MEI Core 1 Coordinate Geometry Section 2: Curves and circles Solutions to Exercise 1. (i)
( x 0)2 ( y 0)2 6 2
x 2 y 2 36 (ii)
( x 3)2 ( y 1)2 5 2
x 2 6 x 9 y 2 2 y 1 25 x 2 y 2 6 x 2 y 15 (iii) ( x 2)2 ( y 5 )2 1 2
x 2 4 x 4 y 2 10 y 25 1 x 2 y 2 4 x 10 y 28 (iv)
( x 0)2 ( y 4)2 3 2
x 2 y 2 8 y 16 9 x 2 y 2 8 y 7
2. (i)
x 2 y 2 100 102 Centre = (0, 0), radius = 10.
x 2 2 y 7
2
(ii)
16 42 Centre = (2, 7), radius = 4
(iii)
x 3 2 y 4
4 22 Centre = (-3, 4), radius = 2
(iv)
x 4 2 y 5
2
2
20
Centre = (-4, -5), radius 20
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MEI C1 Coord. geom. Section 2 Exercise 3. (i)
x 2 y 2 4x 5 0 x 2 4x y 2 5 0 ( x 2)2 4 y 2 5 0 ( x 2)2 y 2 9 Centre = (-2, 0), radius = 3.
(ii)
x 2 y 2 6 x 10 y 20 0 x 2 6 x y 2 10 y 20 0 ( x 3)2 9 ( y 5 )2 25 20 0 ( x 3)2 ( y 5 )2 14 Centre is (3, -5) and radius 14
(iii)
x 2 y 2 2 x 3y 3 0 x 2 2 x y 2 3y 3 0 ( x 1)2 1 ( y 23 )2
9 4
30
( x 1)2 ( y 23 )2 1
9 4
3
( x 1) ( y ) 2
3 2 2
1 4
Centre is (1, 23 ) and radius 21 .
4. Radius of circle (6 4)2 (3 ( 2))2 4 25 29 Equation of circle is ( x 4)2 ( y 2)2 29
x 2 8 x 16 y 2 4 y 4 29 x 2 y 2 8 x 4y 9
5. Centre of circle C is the midpoint of AB. 2 6 04 , C (4, 2) 2 2 Radius of circle is distance AC (2 4)2 (0 2)2 4 4 8 Equation of circle is ( x 4)2 ( y 2)2 8
x 2 8 x 16 y 2 4 y 4 8 x 2 y 2 8 x 4 y 12 0
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MEI C1 Coord. geom. Section 2 Exercise 6.
R (0, 9) M
C (a , b )
Q (0, 3)
The midpoint M of QR is (0, 6). Since a diameter which passes through M is perpendicular to QR, then the line CM must be horizontal, and therefore b = 6. Since the circle touches the x-axis, the radius of the circle must be b, i.e. 6. The equation of the circle is therefore ( x a )2 ( y 6)2 62 The circle passes through (0, 3), so (0 a )2 (3 6)2 6 2
a 2 9 36 a 2 27 a 27 3 3 The equation of the circle is either ( x 3 3)2 ( y 6)2 36 or ( x 3 3)2 ( y 6)2 36 .
7. (i)
x 2 y2 8 Substituting in y 4 x gives x 2 (4 x )2 8
x 2 16 8 x x 2 8 2x2 8x 8 0
x 2 4x 4 0 ( x 2)2 0 The line meets the circle at just one point, so the line touches the circle and is therefore a tangent. (ii)
x 2 y 2 25 Substituting in 4 y 3 x 25 gives
y
3 x 25 4
x 2 y 2 25
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MEI C1 Coord. geom. Section 2 Exercise 2
3 x 25 x 25 4 3 x 25 2 x2 25 16 16 x 2 9 x 2 150 x 625 400 2
25 x 2 150 x 225 0
x 2 6x 9 0 ( x 3)2 0 The line meets the circle at just one point, so the line touches the circle and is therefore a tangent.
8. (i) Substituting y 3 x 1 into y x 2 4 x 7 : 3x 1 x 2 4x 7
x2 7x 6 0 ( x 1)( x 6) 0 x 1 or x 6 When x 1 , y 3 1 1 4 When x 6 , y 3 6 1 19 The intersection points are (1, 4) and (6, 19). (ii) Substituting y x 2 into y x 2 2 x 8 :
x 2 x2 2x 8 x2 x 6 0 ( x 3)( x 2) 0 x 3 or x 2 When x 3 , y 3 2 5 When x 2 , y 2 2 0 The intersection points are (-3, -5) and (2, 0).
9. x 2 y 2 65 2 y x 10 x 10 2 y Substituting in: (10 2 y )2 y 2 65
100 40 y 4 y 2 y 2 65 5 y 2 40 y 35 0
y2 8y 7 0 ( y 1)( y 7) 0 y 1 or y 7 When y = 1, x 10 2 1 8
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MEI C1 Coord. geom. Section 2 Exercise When y = 7, x 10 2 7 4 so P is (8, 1) and Q is (-4, 7) Length PQ (8 ( 4))2 (1 7)2 144 36 180
7 6 1 5 ( 2) 7 7 0 7 Gradient of QR 7 5 6 1
10. (i) Gradient of PR
Gradient of PR × gradient of QR
1 7 1 7
so PR and QR are perpendicular. (ii) The angle in a semicircle is 90°, so PQ must be a diameter. (iii) Since PQ is a diameter, the centre C of the circle is the midpoint of PQ. 2 6 6 0 C , (2, 3) 2 2 Radius of circle = length CQ (6 2)2 (0 3)2
16 9 25 5 Equation of circle is ( x 2)2 ( y 3)2 25 .
11. (i)
x 2 y 2 17
(ii) Substituting x = -4 and y = -1: x 2 y 2 ( 4)2 ( 1)2 16 1 17
1 0 1 4 0 4 Tangent to circle at P is perpendicular to radius OP so gradient of tangent = -4 Equation of tangent is y ( 1) 4( x ( 4))
(iii) Gradient of radius OP
y 1 4( x 4) y 1 4 x 16 y 4 x 17 0 (iv) x y 3 y 3 x Substituting into equation of circle: x 2 (3 x )2 17
x 2 9 6 x x 2 17 2 x 2 6x 8 0
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MEI C1 Coord. geom. Section 2 Exercise x 2 3x 4 0 ( x 4)( x 1) 0 x 4 or x 1 When x = 4, y 3 4 1 When x = -1, y 3 ( 1) 4 Coordinates of Q and R are (4, -1) and (-1, 4). (v) Tangent is y 4 x 17 0 Substituting in y 3 x gives (3 x ) 4 x 17 0
20 3 x 0
x 203 When x 203 , y 3 203
29 3
Coordinates of S are 203 , 113
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Circles Student sheets Equations of Circles Cut out the equation cards on the first two pages and the circle diagram cards on the following three pages, and match up the equations with the circles.
(x – 3)² + (y + 3)² = 9 (x + 3)² + (y + 3)² = 9 (x – 4)² + (y + 3)² = 9 (x + 4)² + (y + 3)² = 9 (x – 2)² + (y + 3)² = 4 (x + 2)² + (y + 3)² = 4 (x – 3)² + (y – 3)² = 9 (x + 3)² + (y – 3)² = 9 (x – 4)² + (y – 3)² = 9 (x + 4)² + (y – 3)² = 9 (x – 2)² + (y – 3)² = 4 (x + 2)² + (y – 3)² = 4 (x – 3)² + (y + 1)² = 4 (x 3)² + (y 1)² = 4 (x + 1)² + (y – 3)² = 4 (x + 1)² + (y + 3)² = 4 (x – 4)² + (y + 4)² = 8 (x + 4)² + (y + 4)² = 8 (x + 3)² + (y + 1)² = 4 (x + 3)² + (y – 1)² = 4 © Susan Wall, MEI, 2004
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Circles Student sheets
(x – 1)² + (y – 3)² = 4 (x – 1)² + (y + 3)² = 4 (x – 4)² + (y – 4)² = 8 (x + 4)² + (y – 4)² = 8 (x + 2)² + (y – 3)² = 9 (x + 2)² + (y + 3)² = 9 (x – 3)² + (y + 2)² = 9 (x + 3)² + (y + 2)² = 9 (x + 2)² + (y + 3)² = 16 (x – 2)² + (y + 3)² = 16 (x – 2)² + (y – 3)² = 9 (x – 2)² + (y + 3)² = 9 (x – 3)² + (y – 2)² = 9 (x + 3)² + (y – 2)² = 9 (x – 2)² + (y – 3)² = 16 (x + 2)² + (y – 3)² = 16
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Circles Student sheets 1
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5
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12
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Circles Student sheets 13
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Circles Student sheets 25
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Circles Student sheets Which One and why? Problem: To find the equation of the tangent to the circle:
x 2 4 x y 2 10 y 72 , at the point (8, 4) . Cross out the incorrect versions of the solution:
x 2 4 x y 2 10 y
( x 2) 2 4 ( y 5) 2 25 ( x 2) 2 4 ( y 5) 2 25 ( x 2) 2 4 ( y 5) 2 25 ( x 2) 2 4 ( y 5) 2 25
Because:
So x 4 x y 10 y 72 becomes: 2
2
( x 2) 2 ( y 10) 2
43 51 101 93
Because:
© Susan Wall, MEI, 2004
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Circles Student sheets The centre of the circle is:
( 2, 5) ( 2, 5) (2, 5) (2, 5) Because:
The gradient of the line joining the centre to the point (8, 4) is:
1 10
1 10
10 1
6 1
1 6
9 10
10 9
6 9
9 6
9 6
10 1
1 6
6 1
10 9
9 10
6 9
Because:
Therefore the equation of the tangent is:
© Susan Wall, MEI, 2004
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MEI Core 1 Coordinate Geometry Section 2: Curves and circles Multiple Choice Test Do not use a calculator in this test. 1) A circle has the equation x² + y² = 16. The radius of this circle is (a) 8 (c) 16 (e) I don’t know
(b) 4 (d) 256
2) A circle has the equation (x + 3)² + (y − 1)² = 4. Which of the following statements is false? (a) The y coordinate of the centre is −1 (c) The x coordinate of the centre is −3 (e) I don’t know
(b) The radius of the circle is 2 (d) The point (−3,−1) lies on the circle
3) The equation of a circle with centre (2, 1) and radius 6 is (a) (x + 2)² + (y + 1)² = 36 (c) (x − 2)² + (y − 1)² = 6 (e) I don’t know
(b) (x + 2)² + (y + 1)² = 6 (d) (x – 2)² + (y – 1)² = 36
4) The equation of a circle with radius 5 and centre (3, -2) can be written as (a) x 2 + y 2 − 3x + 2 y = 25
(b) x 2 + y 2 + 3x − 2 y = 25
(c) x 2 + y 2 − 6 x + 4 y = 12 (e) I don’t know
(d) x 2 + y 2 + 6 x − 4 y = 12
Questions 5 and 6 are about the circle x 2 + y 2 − 2 x + 6 y = 10 . 5) The centre of the circle is (a) (1, -3) (c) (2, -6) (e) I don’t know
(b) (-1, 3) (d) (-2, 6)
© MEI, 25/07/08
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MEI C1 Coord. geom. Section 2 MC test 6) The radius of the circle is (a) 20 (c) 20 (e) I don’t know
(b) 10 (d) 10
7) O is the centre of a circle. P is a point on the circumference. The gradient of OP is 2. The gradient of the tangent at P is (a) 2 (c) −2 (e) I don’t know
(b) 0.5 (d) −0.5
8) The equation of a line is y = x. The equation of a circle is x² + y² = 8. Which one of the following statements is true? (a) The line does not meet the circle (c) The line cuts the circle at one point (e) I don’t know
(b) The line cuts the circle at two points (d) The line is a tangent to the circle
9) AB is the diameter of a circle centre O. P is a point on the circumference. Which one of the following statements is true? (a) When P is equidistant from A and B then OP is parallel to AB (c) AP² + PB² = AB² (e) I don’t know
(b) Angle APB varies as the position of P varies (d) Triangle APB is acute angled
10) The line y = 2 x + 3 is a tangent to a circle with centre (2, -3). The radius of the circle is (a) 20 (c) 20 (e) I don’t know
(b) 40 (d) 40
© MEI, 25/07/08
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MEI Core 1 Coordinate Geometry Chapter assessment Do not use a calculator in this test. 1. A line l1 has equation 5 y + 4 x = 3 . (i) Find the gradient of the line. [1] (ii) Find the equation of the line l2 which is parallel to l1 and passes through the point (1, -2). [3]
2. Describe fully the curve whose equation is x 2 + y 2 = 4 .
[2]
3. The coordinates of two points are A (-1, -3) and B (5, 7). Calculate the equation of the perpendicular bisector of AB. [4] 4. Show that the line y = 3x – 10 is a tangent to the circle x 2 + y 2 = 10 .
[4]
5. The line y = 2 x − 3 meets the x-axis at the point P, and the line 3 y + 4 x = 8 meets the x-axis at the point Q. The two lines intersect at the point R. (i) Find the coordinates of R. [4] (ii) Find the area of triangle PQR. [3] 6. The equation of a circle is x 2 + y 2 − 4 x + 2 y = 15 (i) Find the coordinates of the centre C of the circle, and the radius of the circle. (ii) Show that the point P (4, -5) lies on the circle. (iii) Find the equation of the tangent to the circle at the point P.
[3] [1] [4]
7. The coordinates of four points are P (-2, -1), Q (6, 3), R (9, 2) and S (1, -2). [4] (i) Calculate the gradients of the lines PQ, QR, RS and SP. [1] (ii) What name is given to the quadrilateral PQRS? [2] (iii) Calculate the length SR. (iv) Show that the equation of SR is 2y = x – 5 and find the equation of the line L through Q perpendicular to SR. [5] [3] (v) Calculate the coordinates of the point T where the line L meets SR. [3] (vi) Calculate the area of the quadrilateral PQRS. 8. AB is the diameter of a circle. A is (1, 3) and B is (7, -1). [2] (i) Find the coordinates of the centre C of the circle. [2] (ii) Find the radius of the circle. [2] (iii) Find the equation of the circle. (iv) The line y + 5x = 8 cuts the circle at A and again at a second point D. Calculate the coordinates of D. [4] [3] (v) Prove that the line AB is perpendicular to the line CD. Total 60 marks
© MEI, 25/07/08
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