Mathematics

Circle - 1 Session

Session Objectives

Session Objective 1. Definition -Locus 2. Different forms of circle -Standard form -End-point diameter form -General form -Parametric form 3. Condition for second degree equation to represent a circle

radius Definition-Locus Definition: locus of a point in a plane which is at a fixed given distance from a fixed given point in the plane is circle. R C(x c,y c ) x y P(x,y) C(x c,y c )  center; Rradius P(x,y)  Any point on the circle center O(0,0)

Definition-Locus By distance formula (x-x c ) 2 +(y-y c ) 2 =R 2 R C(x c,y c ) P(x,y) Locus of any point on the circle PC 2 = This is also called Standard form

. C(x c,y c ) P(x,y). C(h,k) P(x,y)... Standard form Standard form of the circle center at C(x c,y c ) and radius R (x-x c ) 2 +(y-y c ) 2 =R 2 x y When origin coincides with center of circle Corresponding equation is called center at origin form C(0,0) P(x,y) R Special Cases: x 2 +y 2 =r 2 Case1

Illustrative Problem Find the equation of the circle having center at (3,4) and passes through (6,7) Find radius Radius=distance of points (3,4) & (6,7)=18 Use Standard form Ans: (x-3) 2 +(y-4) 2 =18 Solution:

Illustrative Problem Find the equation of the circle having x-y=0 and 3x-4y+1=0 as diameters and passes through (2,3) Solution x-y=0 3x-4y+1=0 Point of intersection of x-y=0 and 3x-4y+1=0 x-y=0 3x-4y+1=0 By Solving, we get Center(1,1) Center Of Circle P(2,3) O (1,1) r 2 =PO 2 =(2-1) 2 +(3-1) 2 =5 Equation of circle: (x-1) 2 +(y-1) 2 =5 r

. A(x 1,y 1 ) B(x 2,y 2 ) y x End-Point Diameter Form Equation of the circle whose end- points of one diameter are given as A(x 1,y 1 ) and B(x 2,y 2 ) P(x,y) Slope of AP=(y-y 1 )/(x-x 1 ) Slope of BP=(y-y 2 )/(x-x 2 ) (Slope of AP) x (slope of BP)=-1 APBP Any points on the circle O( 0,0 )

End-Point Diameter Form (y-y 1 )(y-y 2 )= - (x-x 1 )(x-x 2 ) (x-x 1 )(x-x 2 )+(y-y 1 )(y-y 2 )=0. A(x 1, y 1 ) B(x 2,y 2 ) y x P(x,y) Any points on the circle End-point diameter form of circle O( 0,0 )

Illustrative Problem Find the centre and radius of the circle represented by (x-x 1 )(x-x 2 )+(y-y 1 )(y-y 2 )=0 Two extremities of diameter  P(x 1,y 1 ) and Q(x 2,y 2 ) Centre is the mid-point of PQ Radius=CP=CQ.C P(x 1,y 1 ) Q(x 2,y 2 ) Solution:

Illustrative Problem Find the equation of circle of diameter 5 units, passes through (4,0) and has 3x+4y-12=0 as equation of diameter Solution: Observe:(4,0) lies on 3x+4y-12=0 3x+4y-12=0 Equation of the line is x=4+r cos  y= 0+r sin  Parametric form of line tan =slope of line=(-3/4)  cos =(-4/5) and sin =(3/5) (4,0)

Solution Cont. Let the other end of diameter be (x 1,y 1 ) r=  5 and cos =(-4/5) and sin =(3/5) x 1 =4+5 (-4/5)=0 y 1 =0+ 5 (3/5)=3 or x 1 =4-5 (-4/5) =8 y 1 = 0-5 (3/5) =-3 Other end of the diameter is either(0,3) or (8,-3) for r=5 for r=-5

3x+4y-12=0 (4,0) (0,3) (8,-3) (x-0)(x-4)+(y-3)(y-0)=0 or x(x-4)+y(y-3)=0 (x-8)(x-4)+(y+3)(y-0)=0  x 2 +y 2 -4x-3y=0 x 2 +y 2 -12x+3y+32=0 Solution Cont.

General Form Standard form: (x-x c ) 2 +(y-y c ) 2 =R 2 ………….(1)  x 2 +y 2 -2x c x-2y c y+ x c 2 + y c 2 - R 2 =0 In general form it is written as : x 2 +y 2 +2gx+2fy+ c=0, where g, f, c R Comparing with (1) g=-x c, f=-y c c= x c 2 + y c 2 - R 2

General Form x 2 +y 2 +2gx+2fy+ c=0 represents circle having center at (-g,-f) general form of circle 1. g 2 +f 2 -c>0 real circle 2. g 2 +f 2 -c=0  3. g 2 +f 2 -c<0  (R>0) ? Point circle (R=0) ? Imaginary circle (R<0) and radius=

Illustrative Problem Find the centre & radius of the circle 2x 2 +2y 2 -6x+6y-5=0 Make co-eff of x 2 & y 2 as 1 x 2 +y 2 -3x+3y-(5/2)=0 g=-(3/2); f=3/2;c=-5/2 Centre(3/2,-3/2) Centre (3,- 3)? Solution: Note:- center is (-g,-f) when co-eff of x 2 & y 2 is 1

Illustrative Problem Show that the four points (1,0), (2,-7),(8,1) and (9,-6) are concylic. i.e,lie on the same circle. Solution: Let x 2 +y 2 +2gx+2fy+c=0 be the equation of the circle. As it passes through (1,0)  g.1+2f.0+c=0  2g+c= (i) As it passes through (2,-7) 2 2 +(-7) 2 +2g.2+2f.(-7)+c=0 4g-14f+c= (ii) As it passes through (8,1)  g.8+2f.1+c=0 16g+2f+c= (iii)

Solution Cont. 2g+c= (i) 4g-14f+c= (ii) 16g+2f+c= (iii) (ii)-(i)  2g-14f=-52 (ii)-(iii)  -12g-16f=12 or, g-7f = -26 or, 3g+4f = -3 By Solving g=-5,f=3  c=9 Equation of the circle is: x 2 +y 2 -10x+6y+9=0 For (9,-6) :L.H.S= 9 2 +(-6) (-6)+9 = 0 Four points are Concylic

Condition For 2 nd Degree Equ n to Represent Circle General form of 2 nd degree equn:- ax 2 +by 2 +2gx+2fy+2hxy+c=0---(1) where a,b,g,f,h,c  R It represents curves of different kind AS : 1. for a=b=h=0 (and g,f  0) 2gx+2fy+c=0  equ n (1) be straight line

General equation of the circle : x 2 +y 2 +2gx+2fy+c= (ii) So general 2 nd degree equation will represent circle iff 1. Co-eff of x 2 =Co-eff y 2 0 2. Co-eff of xy=0 ax 2 +by 2 +2gx+2fy+2hxy+c=0---(i) General 2 nd degree equation: Condition For 2 nd Degree Equ n to Represent Circle  a = b  0  h=0 By Comparing (i) and (ii)

Illustrative Problem Given that ax 2 +by 2 +2gx+2fy+c=0 represents circle. Then find the radius of the circle? Solution: As ax 2 +by 2 +2gx+2fy+c=0---(1) represents a circle a=b  0 Equation (1) can be written as

Illustrative problem Find the equation of the circle passes through the point of intersection of lines x+y=6, 2x+y=4 and x+2y=5 Solution: Let L 1  x+y-6=0 L 2  2x+y-4=0 L 3  x+2y-5=0 L 1 =0 L 2 =0 L 3 =0 p1p1 p2p2 p3p3 L 1 L 2 + L 1 L 3 +  L 3 L 2 =0 What does it represent?

Solution Cont. L 1 =0 L 2 =0 L 3 =0 p1p1 p2p2 p3p3 f(x,y)=L 1 L 2 + L 1 L 3 +  L 3 L 2 f(x,y)= (x+y-6)(2x+y-4)+(x+y-6)(x+2y-5) +(x+2y-5)(2x+y-4) (1) (x 1,y 1 ) (x 2,y 2 ) (x 3,y 3 ) As P 1 (x 1,y 1 ) lies on L 1 =0 and L 2 =0 x 1 +y 1 -6=0 and 2x 1 +y 1 -4=0 By putting P 1 (x 1,y 1 ) in (1)

Solution Cont. L 1 =0 L 2 =0 L 3 =0 p1p1 p2p2 p3p3 (x 1,y 1 ) (x 2,y 2 ) (x 3,y 3 ) f(x 1,y 1 )= (x 1 +y 1 -6)(2x 1 +y 1 -4)+(x 1 +y 1 -6)(x 1 +2y 1 -5) +(x 1 +2y 1 -5)(2x 1 +y 1 -4) x 1 +y 1 -6=0 and 2x 1 +y 1 -4=0 = 0x0+ x0x(x 1 +2y 1 -5) +(x 1 +2y 1 -5)x0  P 1 lies on f(x,y) Similarly it can be proved that P 2 and P 3 also lies on f(x,y). = 0

Solution Cont. P 1,P 2,P 3 lies on f(x,y)=0 points of intersections of L 1 =0, L 2 =0 and L 3 =0 lies on curve f(x,y)=0 L 1 L 2 + L 1 L 3 +  L 3 L 2 =0 Represents curve, which passes through the point of intersection of L 1 =0, L 2 =0 and L 3 =0 L 1 =0 L 2 =0 L 3 =0 p1p1 p2p2 p3p3 (x 1,y 1 ) (x 2,y 2 ) (x 3,y 3 )

Solution Cont. L 1 L 2 + L 1 L 3 +  L 3 L 2 =0 (x+y-6)(2x+y-4) +(x+y-6) (x+2y-5) +(x+2y-5)(2x+y-4)=0 Will represent circle 1.Co-efficient of x 2 =co-efficient of y =1+2+2 (1) 2.Co-efficient of xy= = (2) From (1) and (2): =1, = -6/5

Solution Cont. L 1 L 2 + L 1 L 3 +  L 3 L 2 =0 For =1 and =-6/5 the curve represents circle Equation of circle is (x+y-6)(2x+y-4)+(x+y-6)(x+2y-5) -(6/5)(x+2y-5)(2x+y-4)=0 x 2 +y 2 -17x-19y+50 = 0 is the equation of the circle

Parametric Form For the circle (x-x c ) 2 +(y-y c ) 2 =R 2. C(x c,y c ) P(x,y)  R xcxc (R cos) ycyc (R sin) point P(x,y) on circle represented as: x= x c + R cos  x c +Rcos y c +Rsin collectively represents all the points on circle for different values of : 0   360 w.r.t x-axis y= y c + R sin  =x =y

. P(x,y)  R Parametric Form Parametric form (x-x c ) 2 +(y-y c ) 2 = R 2 Special case: For the circle x 2 +y 2 =R 2 Parametric form:- x y R cos R sin

Illustrative Problem Find the parametric equation of the circle x 2 +y 2 -4x-2y+1=0 Find the center(h,k) & radius(r) h=2, k=1, r=2 Put in the parametric form x=(2+2cos); y=(1+2 sin ) Solution:

Class Exercise

Class Exercise -1 Find the equation of the circle whose diameters are 2x–3y+12=0 and x+4y–5=0 and area is 154 Solution: 2x-3y+12=0 x+4y-5=0 Point of intersection of x+4y-5=0 and 2x-3y+12=0 Center Of Circle O (-3,2) 2x-3y+12=0 x+4y-5=0 By Solving, we get Center(-3,2) Let the radius of circle is r. Equation of circle is (x+3) 2 +(y–2) 2 =49

Class Exercise - 2 Find the parametric form of the circle: x 2 +y 2 +px+py=0. Solution: x 2 +y 2 +px+py=0. Parametric Form of Equation

Class Exercise - 3 Find the equation of the circle passing through the vertices of the triangle whose sides are along x+y=2, 3x–4y=6 and x–y=0. Find its center and radius. Solution: Let L 1  x+y-2=0 L 2  3x-4y-6=0 L 3  x-y=0 Equation of circle passes through point of intersection of L 1, L 2 and L 3 is

Solution Cont. Equation (i) will represent circle if (a) co-efficient of x 2 = co-efficient of y 2 (b) co-efficient of xy = 0 from (2):  =-25/7

Solution Cont. Equation of circle is: (x+y–2)(3x–4y–6)–(1/7)(3x–4y–6)(x–y) –(25/7)(x–y)(x+y–2)=0  x 2 +y 2 –4x+14y–12=0

Class Exercise - 4 Find the equation of a circle which passes through origin, cut positive x and positive y-axis at 4 and 6 units from origin respectively Solution: O (0,0) B (4,0) C (0,6) Diameter,asBOC=90 0 A (0, 6), B(4, 0)are the end points of diameter. Equation of circle is: (x–0)(x–4)+(y–0)(y–6)=0 or, x 2 +y 2 –4x–6y=0

Class Exercise - 5 Find the equation of circle, the end- points of whose diameter are the centers of the circles x 2 +y 2 +6x– 14y–1=0andx 2 +y 2 –4x+10y–2=0. Solution:. C 1 (-3,7) S 1 =0. C 2 (2,-5) S 2 =0 Diameter (x+3)(x–2)+(y–7)(y+5)=0 or, x 2 +y 2 +x–2y–41=0

Class Exercise - 6 Find the equation of a circle passes through the points(2,3),(0,–1) and center lies on the line 3x–4y+1=0. Solution: Let the equation of the circle be x 2 +y 2 +2gx+2fy+c=0...(i) Centre : (–g,–f) -3g+4f+1=0...(ii) As(2,3) lies on (i)  13+4g+6f+c= 0...(iii) As(0,–1) lies on (i)  1–2f+c=0...(iv) Putting g and f in (iii)  c=–3; f=–1; g=–1

Solution Cont. x 2 +y 2 +2gx+2fy+c=0…..(i) c=–3; f=–1; g=–1 Equation of the circle is: x 2 +y 2 –2x–2y–3=0

. Class Exercise - 7 Find the equation of the circle which passes through (1, 1), (2, 2) and radius is 1. Solution: A (1,1) B (2,2) C( h,k ) As it passes through P(1,1)  (h–1) 2 +(k–1) 2 =1 r=1 h 2 +k 2 –2h–2k+1=0...(i) As it passes through Q (2,2)  (h–2) 2 +(k–2) 2 =1 h 2 +k 2 –4h–4k+7=0...(ii) (i)—(ii) h+k=3  k=3–h

Solution Cont. h 2 +k 2 –2h–2k+1=0...(i) h 2 +k 2 –4h–4k+7=0...(ii) k=3–h From (i): h 2 +(3–h) 2 –2h–2(3–h)+1=0  2h 2 –6h+4=0 h=1,2 for h=1,k=2;for h=2,k=1 Equation of the circle is (x–1) 2 +(y–2) 2 =1 (x–2) 2 +(y–1) 2 =1 h 2 –3h+2=0

Class Exercise - 8 If one end of diameter of circle x 2 +y 2 –4x–6y+11=0 is (8,4). Find the co-ordinate of other end? Solution: Center of the circle is (2,3) Let the other end be (a,b) other end is (–4, 2)

Class Exercise - 9 Find the equation of circle touching x+y=2 and 2x+2y=3 and passing through (1,1)

Solution x+y=2 2x+2y=3P( 1,1 ) Q Observe: 1. x+y=2 and 2x+2y=3 are the parallel lines. Diameter of the circle= Distance between the parallel lines. 2. Point (1,1) lies on x+y=2 Line through P and perpendicular to the parallel line is diameter of circle It intersect 2x+2y=3 at diametrically opposite point.

Solution Cont. x+y=2 2x+2y=3 P(1,1) Q Slope of the parallel lines =–1 PQ is the diameter of the circle,which is r to the given lines. Slope of PQ = 1  y = x Q intersection of x=y 2x+2y=3 x=y

Solution Cont. x+y=2 2x+2y=3 P(1,1) Q End-Points of the diameter of the circles are P(1,1) and Q(3/4,3/4) Equation of the circle is (x–1)(x–3/4)+(y–1)(y–3/4)=0  (x–1)(4x–3)+(y–1)(4y–3) = 0  4x 2 –7x+4y 2 –7y+6=0

Class Exercise - 10 If the equation of a circle is ax 2 +(2a–3)y 2 –4x–1=0 then the centre is (a)(2,0) (b)(2/3,0) (c) (-2/3,0) (d) None of these Solution: If ax 2 +(2a–3)y 2 –4x–1=0 represents circle a=2a–3  a=3 Equation of circle is: 3x 2 +3y 2 –4x–1=0  x 2 +y 2 –(4/3)x–1=0 Center: (2/3,0)

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