CIRCLES Notice the path in which the tip of the hand of a watch moves. (see Fig..) 0 9 3 8 4 7 6 5 Fig.. Fig.. Again, notice the curve traced out when a nail is fied at a point and a thread of certain length is tied to it in such a wa that it can rotate about it, and on the other end of the thread a pencil is tied. Then move the pencil around the fied nail keeping the thread in a stretched position (See Fig.) Certainl, the curves traced out in the above eamples are of the same shape and this tpe of curve is known as a circle. The distance between the tip of the pencil and the point, where the nail is fied is known as the radius of the circle. We shall discuss about the curve traced out in the above eamples in more details. OBJECTIVES After studing this lesson, ou will be able to : derive and find the equation of a circle with a given centre and radius; MATHEMATICS 379
Circles state the conditions under which the general equation of second degree in two variables represents a circle; derive and find the centre and radius of a circle whose equation is given in general form; find the equation of a circle passing through : (i) three non-collinear points (ii) two given points and touching an of the aes; derive the equation of a circle in the diameter form; find the equation of a circle when the end points of an of its diameter are given; and find the parametric representation of a circle with given centre and radius. EXPECTED BACKGROUND KNOWLEDGE Terms and concepts connected with circle. Distance between two points with given coordinates. Equation of a straight line in different forms.. DEFINITION OF THE CIRCLE A circle is the locus of a point which moves in a plane in such a wa that its distance from a fied point in the same plane remains constant. The fied point is called the centre of the circle and the constant distance is called the radius of the circle.. EQUATION OF A CIRCLE Can we find a mathematical epression for a given circle? Let us tr to find the equation of a circle under various given conditions... WHEN COORDINATES OF THE CENTRE AND RADIUS ARE GIVEN Let C be the centre and a be the radius of the circle. s of the centre are given to be (h, k), sa. Take an point P(, ) on the circle and draw perpendiculars CM and PN on OX. Again, draw CL perpendicular to PN. We have ( h,) k C P(,) L CL = MN = ON OM = h and PL = PN LN = PN CM = k In the right angled triangle CLP, CL + PL = CP ( = a h) + ( k)...() O M N Fig..3 380 MATHEMATICS
This is the required equation of the circle under given conditions. This form of the circle is known as standard form of the circle. Conversel, if (, ) is an point in the plane satisfing (), then it is at a distance a from (h, k). So it is on the circle. What happens when the (i) circle passes through the origin? (ii) circle does not pass through origin and the centre lies on the -ais? (iii) circle passes through origin and the -ais is a diameter? (iv) centre of the circle is origin? (v) circle touches the -ais? (vi) circle touches the -ais? (vii) circle touches both the aes? We shall tr to find the answer of the above questions one b one. (i) In this case, since (0, 0) satisfies (), we get h + k = a Hence the equation () reduces to + h k...() (ii) In this case k Hence the equation () reduces to ( = a h) +...(3) ' C O ' O C ' ' Fig..4 MATHEMATICS 38
(iii) In this case k and h = ± a (see Fig..4) Hence the equation () reduces to + ± a...(4) Circles (iv) In this case h = k Hence the equation () reduces to + = a...(5) (v) In this case k = a (see Fig..5) Hence the equation () reduces to + h a + h...(6) C a ' O ' h Fig..5 (vi) In this case h = a Hence the equation () reduces to + a k + k...(7) (vii) In this case h = k = a. (See Fig..6) Hence the equation () reduces to + a a + a...(8) C ' O ' Fig..6 38 MATHEMATICS
Eample. Find the equation of the circle whose centre is (3, 4) and radius is 6. Solution : Comparing the terms given in equation (), we have h = 3, k = 4 and a = 6. ( 3) + ( + 4) = or + 6 + 8 6 Eample. Find the centre and radius of the circle given b ( + ) + ( ) = 4. Solution: Comparing the given equation with ( h) + ( k) = a we find that h =, k =, a = 4 h =, k =, a =. So the given circle has its centre (,) and radius..3 GENERAL EQUATION OF THE CIRCLE IN SECOND DEGREE IN TWO VARIABLES The standard equation of a circle with centre (h, k) and radius r is given b ( = r h) + ( k)... () or + h k + h + k r... () This is of the form + + g + f + c. g f c + + + +... (3) ( )( ) + g + g + + f + f = g + f c ()() + g ( ) + + f = g + f c [ ()() g ] [ ] ( ) f g f c + = +... (4) ()() h + k = r where h = g, k = f, r = g + f c This shows that the given equation represents a circle with centre ( g, f) and radius = g + f c MATHEMATICS 383
Circles.3. CONDITONS UNDER WHICH THE GENERAL EQUATION OF SECOND DEGREE IN TWO VARIABLES REPRESENTS A CIRCLE Let the equation be + + g + f + c (i) It is a second degree equation in, in which coefficients of the terms involving and are equal. (ii) It contains no term involving Note : In solving problems, we keep the coefficients of and unit. Eample.3 Find the centre and radius of the circle 45 + 45 60 + 36 + 9 Solution : Given equation can be written on dividing b 45 as 4 4 + + + 3 5 Comparing it with the equation 9 45 g f c + + + + we get 9 g =, f = and c = 3 5 45 Thus, the centre is, 3 and radius is 5 g + f c = 4 5 Eample.4 Find the equation of the circle which passes through the points (, 0), (0, 6) and (3, 4). Solution: Let the equation of the circle be g f c + + + +...() Since the circle passes through three given points so the will satisf the equation (). Hence and + g + c () 36 f + c (3) 5 + 6g + 8 f + c (4) Subtracting () from (3) and (3) from (4), we have and g + f = 35 6g + 0 f = 384 MATHEMATICS
7 Solving these equations for g and f, we get g =, f = 4 69 Substituting g in (), we get c = 47 8 and substituting g, f and c in (), the required equation of the circle is 4 + 4 4 + 47 + 38 Emaple. 5 Find the equation of the circles which touches the ais of and passes through the points (, ) and (3, 4). Solution : Since the circle touches the -ais, put k = a in the standard form (See result 6) of the equation of the circle, we have + h a + h... () This circle passes through the point (, ) h h + 4a + 5... () Also, the circle passes through the point (3, 4) h 6h + 8a + 5...(3) Eliminationg a from () and (3), we get h + h = 5 0 h = 3 or h = 5. From (3) the corresponding values ofa are and 0 respectivel. On substituting the values of h and a in () we get + 6 + 4 + 9... (4) and + + 0 + 0 + 5... (5) (4) and (5) represent the required equations..4 EQUATION OF A CIRCLE WHEN END POINTS OF ONE OF ITS DIAMETERS ARE GIVEN. Let A, ) and B, ) be the given end points of the diameter AB (See Fig..7) ( ( Let P(, ) be an point on the circle drawn on AB as diameter. Join AP and BP. Since the angle in a semi-circle is a right angle. MATHEMATICS 385
AP BP (slope of AP) (slope of BP) =. Now, the slope of AP = and the slope of BP = = ' O ' (), B Fig..7 A P(), (), or ( )( ) + ( )( ) 0...() = Since () is true for ever point on the circle, and for no other point in the plane: () represents the equation of the circle in diameter from. Eample.6 Find the equation of the circle described on the line joining the origin and the point (, 4) as diameter. Solutoon: Here =, ; =, = 4 0 Using the Equation () required the equation of the circle is, ( 0)( ) ( 0( )[ 4) 0 ] + = or + + 4 or + + 4 Eample.7 The equation of a chord of the circle + a is = m. Find the equation of the circle described on this chord as diameter. Solution: The coordinates of the points of intersection of the circle and the given chord are, a ma (0, 0) and, + m + m. Now, for the required equation of the circle these points are the end points of the diameter, so b equation () a 0) + m ma + ( 0) + m ( = 0 386 MATHEMATICS
a ma or + + m + m or ( + m ) + ( + m ) a ma This is the required equation of the circle..5 EQUATION OFA CIRCLE WHEN RADIUS AND ITS INCLINATION ARE GIVEN (PARAMATRIC FORM) In order to find the equation of the circle whose centre is the origin and whose radius is r. Let P(, ) be an point on the circle. Draw PM OX OM =, MP =. Join OP. Let XOP = θ and OP = r Now, = OM = r cos and = MP = r sin Fig..8 Hence the two equations = r cos and = r sin taken together represent a circle. These are known as parametric form of the equations of the circle, where is a parameter. Note : If the centre of the circle is at h and k, then the parametric form of the equation of the circle is = h + r cos and = k + r sin where is a parametre which lies in the interval [0, π]. Eample.8 Find the parametric form of each of the following circles: (i) ( ) + ( + ) = 9 (ii) ( + )( 4)+ ( 3) ( + ) Solution : (i) The equation of the circle is ( ) + ( ) = 3 Comparing it with the standard equation of the circle, we get centre (, ) and radius = 3 for the given circle. Parametric form of the equation of the circle is = +3 cos ; = + 3 sin (ii) Equation of the circle can be written as + MATHEMATICS 387
( a) c e n t r e i s ( 0, 0) a n d r a d i u s i s 3. ( b) c e n t r e i s ( Comparing it with the general equation of the circle and finding centre and radius, we get Centre (, ) and radius = 3 Circles Parametric form of the equation of the circle is = + 3 cos θ, = + 3 sin θ CHECK YOUR PROGRESS.. Find the equation of the circle whose. Find the centre and radius of the circle (a) + + 3 = 6 (b) 4 + 4 + 3 6,3) and radius is 4. 3. Find the equation of the circle which passes through the points (0, ) (, 0) and (0, 0). 4. Find the equation of the circle which touches the -ais and passes through the points (, )and (,) 5. Find the equation of the circle described on the line joining the points (, 3) and (, 6) as diameter. 6. Find the parametric form of the equation of each of the following circles: (a) ( + ) + ( + ) = 4 (b) 4 + 4 + + 3 (c) ( )( + ) + ( )( + ) LET US SUM UP Standard form of the circle ( h) + ( k) = a Centre is (h, k) and radius is a The general form of the circle is + + g + f + c Its centre: ( g, f) and radius = g + f c Diameter form of the circle If the end points of a diameter are (, ) and (, ),then the equation of the circle is ( )( ) + ( )( ) 388 MATHEMATICS
Parametric form of the circle = acosθ, = a sin θ represents the parametric equation of a circle whose centre is at (0, 0) and radius =a If the centre of the circle is at (h, k) then the parametric equation of the circle is = h + a cos θ and = k + a sin θ. SUPPORTIVE WEB SITES http://www.wikipedia.org http://mathworld.wolfram.com TERMINAL EXERCISE. Find the equation of a circle with centre (4, 6) and radius 7.. Find the centre and radius of the circle + + 4 6 3. Find the equation of the circle passes through the point (,0), (,0) and (0,) 4. Find the parametric form of the equation of circles given below : (a) + = 3 (b) + 4 + 6 = MATHEMATICS 389
ANSWERS CHECK YOUR PROGRESS.. (a) + = 9 (b) + + 4 6 3 Circles. (a) 3, ; 37 (b) 3, ; 4 8 09 8 3. + 4. + + + 5. + 9 + 4 6. (a) = + cos θ and = + sin θ. (b) 7 = + cosθ 4 8 7 = + sin θ 4 8 (c) = cos θ and = sin θ. TERMINAL EXERCISE. + 8 + + 3. Centre (, 3); Radius = 3 3. + =. 4. (a) = 3 cos θ, = 3 sin θ. (b) = + 5cos θ, = + 5sin θ. 390 MATHEMATICS