# 2.6. The Circle. Introduction. Prerequisites. Learning Outcomes

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1 The Circle 2.6 Introduction A circle is one of the most familiar geometrical figures. In this brief Section we discuss the basic coordinate geometr of a circle - in particular the basic equation representing a circle in terms of its centre and radius. Prerequisites Before starting this Section ou should... Learning Outcomes After completing this Section ou should be able to... 1 understand what is meant b a function and use functional notation 2 be able to plot graphs of functions write the equation of an given circle obtain the centre and radius of a circle from its equation

2 1. Equations for Circles in the 0 plane The obvious characteristic of a circle is that ever point on its circumference is the same distance from the centre. This fied distance is called the radius of the circle and is generall denoted b R or r or a. In coordinate geometr terms suppose (, ) denotes the coordinates of a point e.g. (4,2) means =4,=2,( 1, 1) means = 1, =1and so on. See Figure 1. ( 1, 1) (4, 2) Figure 1 Write down the distances d 1 and d 2 from the origin of the points with coordinates (4,2) and ( 1, 1) respectivel. Generalise our results to obtain the distance d from the origin of an arbitrar point with coordinates (, ). Using Pthagoras Theorem: (a) d 1 = = 20 is the distance between the origin (0,0) and the point (4,2). (b) d 2 = ( 1) = 2isthe distance between the origin and ( 1, 1). (c) d = is the distance from the origin to an arbitrar point (, ). Note that the positive square root is taken in each case. HELM (VERSION 1: March 18, 2004): Workbook Level 0 2

3 Circles with centre at the origin Suppose (, ) isanpoint P on a circle of radius R whose centre is at the origin. See Figure 2. (, ) P R Figure 2 Bearing in mind the final result of the previous guided eercise, write down an equation relating, and R. Since is distance of an point (, ) from the origin, then for an point P on the above circle = R or = R 2 The whole point is that as the point P in Figure 2 moves around the circle its and coordinates change. However P will remain at the same distance R from the origin b the ver definition of a circle. Hence we sa that = R or, more usuall, = R 2 (1) is the equation of the circle radius R centre at the origin. What this means is that if a point (, ) satisfies (1) then it lies on the circumference of the circle radius R. If (, ) doesnot satisf (1) then it does not lie on that circumference. Note carefull that the right hand sides of the circle equation (1) is the square of the radius. 3 HELM (VERSION 1: March 18, 2004): Workbook Level 0

4 Consider the circle centre at the origin and of radius 5. (a) Write down the equation of this circle. (b) Find which of the following points lie on the circumference of this circle. (5, 0) (0, 5) (4, 3) ( 3, 4) (2, 21) ( 2 6, 1) (1, 4) (4, 4) (c) For those points in (ii) which do not lie on the circle deduce whether the lie inside or outside the circle. (a) =5 2 =25isthe equation of the circle. (b) For each point (, ) wecalculate If this equals 25 the point lies on the circle. (c), on circle (0, 5) on circle on circle ( 3, 4) on circle (2, 21) on circle ( 2 6, 1) on circle inside circle (4, 4) outside circle conclusion (5,0) (4,3) (1,4) Figure 3 demonstrates some of the results of the previous eample. HELM (VERSION 1: March 18, 2004): Workbook Level 0 4

5 2 + 2 =25 (1, 4) (4, 3) (5, 0) ( 3, 4) (0, 5) (4, 4) Figure 3 Note that the circle centre at the origin and of radius 1 has a special name the unit circle. (a) Calculate the distance between the points P 1 ( 1, 1) and P 2 (4, 2). Refer back to Figure 1. (b) Generalise our result to obtain the distance between an two points whose coordinates are ( 1, 1 ) and ( 2, 2 ). 5 HELM (VERSION 1: March 18, 2004): Workbook Level 0

6 P d 2 P 1 A Figure 4 Using Pthagoras Theorem the distance between the two given points is d = (P 1 A) 2 +(AP 2 ) 2 where P 1 A =4 ( 1) = 5, AP 2 =2 1=1 d = = 26 (ii) ( 1, 1 ) P 1 d P 2 A ( 2, 2 ) Figure 5 Between the arbitrar points P 1 and P 2 the distance is d = (AP 2 ) 2 +(P 1 A) 2 where AP 2 = 2 1, P 1 A = 1 2 = ( 2 1 ) so d = ( 2 1 ) 2 +( 2 1 ) 2 Remembering the result of part (ii) of this eample we now consider a circle centre at the point C( 0 ; 0 ) and of radius R. Suppose P is an arbitrar point on this circle which has co-ordinates (, ) (i) R P (, ) C ( 0, 0 ) Clearl R = CP = ( 0 ) 2 +( 0 ) 2 Hence, squaring both sides, Figure 6 HELM (VERSION 1: March 18, 2004): Workbook Level 0 6

7 ( 0 ) 2 +( 0 ) 2 = R 2 (2) which is said to be the equation of the circle centre ( 0, 0 ) radius R. Note that if 0 = 0 =0(i.e. circle centre is at origin) then (2) reduces to (1) so the latter is simpl a special case. The interpretation of (2) is similar to that of (1): an point (, ) satisfing (2) lies on the circumference of the circle. Eample The equation ( 3) 2 +( 4) 2 =4 represents a circle of radius 2 (the positive square root of 4) and has centre C (3,4). N.B. There is no need to epand the terms on the left-hand side here. The given form reveals quite plainl the radius and centre of the circle. Write down the equations of each of the following circles. The centre C and radius R are given: (a) C(0, 2), R=2 (b) C( 2, 0), R=3 (c) C( 3, 4), R=5 (d) C(1, 1), R= 3 7 HELM (VERSION 1: March 18, 2004): Workbook Level 0

8 (a) 0 =0, 0 =2,R 2 =4sob (2) 2 +( 2) 2 =4 is the required equation. (b) 0 = 2, 0 =0,R 2 =9 ( +2) =9 (c) ( +3) 2 +( 4) 2 =25 (d) ( 1) 2 +( 1) 2 =3 Again we emphasize that the right hand side of each of these equations is the square of the radius. Write down the equations of each of the circles shown (a) 2 (b) (c) 3 (d) Figure 7 HELM (VERSION 1: March 18, 2004): Workbook Level 0 8

9 (a) =4 (centre origin, radius 2) (b) ( 1) =1 (centre (1,0) radius 1) (c) ( 3) 2 +( 3) 2 =0 (centre (3,3) radius 3) (d) ( +1) 2 +( +1) 2 =1 (centre ( 1, 1) radius 1) Consider again the equation of the circle, centre (3,4) of radius 2: ( 3) 2 +( 4) 2 =4 (3) In this form of the equation the centre and radius of the circle can be clearl identified and, as we said, there is no advantage in squaring out. However, if we did square out the equation would become =4 or =0 (4) Equation (4) is of course a valid equation for this circle but, we cannot immediatel obtain the centre and radius from it. For the case of the general circle of radius R ( 0 ) 2 +( 0 ) 2 = R 2 epand out the square terms and simplif. It follows from the above eercise that an equation of the form where c = R c = R 2 =0 or We obtain g 2f + c =0 (5) represents a circle with centre (g, f) and a radius obtained b solving for R. Thus c = g 2 + f 2 R 2 9 HELM (VERSION 1: March 18, 2004): Workbook Level 0

10 R = g 2 + f 2 c There is no reason to remember equation (6). In an specific problem the technique of completion of square can be used to turn an equation of the form (5) into the form (2) and hence obtain the centre and radius of the circle. NB. The ke point about (5) is that the coefficients of the term 2 and 2 are the same, i.e. 1. An equation with the coefficient of 2 and 2 identical with value k 1could be converted into the form (5) b division of the whole equation b k. (6) If =0 obtain the centre and radius of the circle that this equation represents. Begin b completing the square separatel on the terms and the terms. 2 2 = ( 1) = ( +5) 2 25 Now complete the problem. The original equation =0 becomes ( 1) 2 1+( +5) =0 ( 1) 2 +( +5) 2 =10 which represents a circle with centre (1, 5) and radius 10. HELM (VERSION 1: March 18, 2004): Workbook Level 0 10

11 Circles and functions Let us return to the equation of the unit circle =1 Solving for we obtain = ± 1 2. This equation does not represent a function because of the two possible square roots which impl that for an value of there are two values of. (You will recall from earlier in this Workbook that a function requires onl one value of the dependent variable corresponding to each value of the independent variable.) However two functions can be obtained in this case: = 1 =+ 1 2 = 2 = 1 2 whose graphs are the semicircles shown. = = 1 2 Figure 8 Annuli between circles Equations in and, such as (1) and (2) for circles, define curves in the O plane. However, inequalities are necessar to define regions. For eample, the inequalit < 1 is satisfied b, all points inside the unit circle - for eample (0, 0) (0, 1 2 )(1 4, 0) ( 1 2, 1 2 ). Similarl > 1 is satisfied b all points outside that circle such as (1,1) > = < HELM (VERSION 1: March 18, 2004): Workbook Level 0

12 Figure 9 Sketch the regions in the O plane defined b (a) ( 1) < 1 (b) ( 1) > 1 The equalit ( 1) =1 (7) is satisfied b an point on the circumference of the circle centre (1,0) radius 1. Then, remembering that ( 1) is the square of the distance between an point (, ) and (1,0), it follows that (a) ( 1) < 1issatisfied b an point inside this circle (Region (A) on the following diagram.) (b) ( 1) > 1 defines the region eterior to the circle since this inequalit is satisfied b ever point outside. (Region (B) on the diagram.) (B) (A) (B) Figure 10 HELM (VERSION 1: March 18, 2004): Workbook Level 0 12

13 The region between two circles with the same centre (concentric circles) is called an annulus or annular region. An annulus is defined b two inequalities. For eample the inequalit > 1 (8) defines, as we saw, the region outside the unit circle. The inequalit < 4 (9) defines the region inside the circle centre origin radius 2. Hence points (, ) which satisf both the inequalities (8) and (9) lie in the annulus between the two circles. The inequalities (8) and (9) are combined b writing 1 < < 4 1 < < Figure 11 Sketch the annulus defined b the inequalities 1 < ( 1) < 9 (10) (Refer back to the previous but one Guided Eercise if necessar.) 13 HELM (VERSION 1: March 18, 2004): Workbook Level 0

14 The quantit ( 1) is the square of the distance of a point (, ) from the point (1,0). Hence, as we saw earlier, the left-hand inequalit 1 < ( 1) or, equivalentl, ( 1) > 1 is the region eterior to the circle C 1 centre (1,0) radius 1. Similarl the right had inequalit ( 1) < 9 defines the interior of the circle C 2 centre (1,0) radius 3. Hence the inequalit (10) holds for an point in the annulus between C 1 and C 2. C 2 C Figure 12 HELM (VERSION 1: March 18, 2004): Workbook Level 0 14

15 Eercises 1. Write down the radius and the coordinates of the centre of the circle for each of the following equations (a) =16 (b) ( 4) 2 +( 3) 2 =12 (c) ( +3) 2 +( 1) 2 =25 (d) 2 +( +1) 2 4=0 (e) ( +6) =0 2. Obtain in each case the equation of the given circle (a) centre C (0, 0) radius 7 (b) centre C (0, 2) radius 2 (c) centre C (4, 4) radius 4 (d) centre C ( 2, 2) radius 4 (e) centre C ( 6, 0) radius 5 3. Obtain the radius and the coordinates of the centre for each of the following circles (a) =0 (b) =11 (c) =0 4. Describe the regions defined b each of these inequalities (a) > 4 (b) < 16 (c) the inequalities in (i) and (ii) together 5. Write an inequalit that describes the points that lie outside the circle of radius 4 with centre ( 4, 2). 6. Write an inequalit that describes the points that lie inside the circle of radius 6 with centre ( 2, 1). 7. Obtain the equation of the circle which has centre (3, 4) and which passes through the point (0, 5). 8. Show that if A( 1, 1 ) and B( 2, 2 ) are at opposite ends of a diameter of a circle then the equation of the circle is ( 1 )( 2 )+( 1 )( 2 )=0. (Hint: if P is an point on the circle obtain the slopes of the lines AP and BP and recall that the angle in a semi-circle must be a right-angle.) 9. State the equation of the unique circle which touches the ais at the point (2,0) and which passes through the point ( 1, 9). 15 HELM (VERSION 1: March 18, 2004): Workbook Level 0

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