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Mathematics

Mathematics. Session. Circle - 1. 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. y. x.

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Mathematics

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  1. Mathematics

  2. Session Circle - 1

  3. Session Objectives

  4. 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

  5. y x 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 center radius C(xc,yc)  center; Rradius . P(x,y) R P(x,y)  Any point on the circle C(xc,yc) O(0,0)

  6. Definition-Locus By distance formula (x-xc)2+(y-yc)2=R2 PC2= P(x,y) Locus of any point on the circle This is also called Standard form R C(xc,yc)

  7. P(x,y) . C(h,k) P(x,y) . y P(x,y) C(xc,yc) R C(0,0) x Standard form Standard form of the circle center at C(xc,yc) and radius R (x-xc)2+(y-yc)2=R2 Special Cases: Case1 When origin coincides with center of circle . . . Corresponding equation is called center at origin form x2+y2=r2

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

  9. P(2,3) x-y=0 r Point of intersection of x-y=0 and 3x-4y+1=0 Center Of Circle 3x-4y+1=0 Equation of circle: (x-1)2+(y-1)2=5 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 O (1,1) x-y=0 3x-4y+1=0 By Solving, we get Center(1,1) r2=PO2=(2-1)2+(3-1)2=5

  10. y A(x1,y1) . B(x2,y2) x End-Point Diameter Form Equation of the circle whose end-points of one diameter are given as A(x1,y1) and B(x2,y2) Any points on the circle P(x,y) Slope of AP=(y-y1)/(x-x1) Slope of BP=(y-y2)/(x-x2) APBP O(0,0) (Slope of AP) x (slope of BP)=-1

  11. Any points on the circle y A(x1,y1) P(x,y) . B(x2,y2) x End-Point Diameter Form (y-y1)(y-y2)= - (x-x1)(x-x2) (x-x1)(x-x2)+(y-y1)(y-y2)=0 End-point diameter form of circle O(0,0)

  12. Illustrative Problem Find the centre and radius of the circle represented by (x-x1)(x-x2)+(y-y1)(y-y2)=0 Solution: Two extremities of diameter  P(x1,y1) and Q(x2,y2) Centre is the mid-point of PQ Q(x2,y2) .C Radius=CP=CQ P(x1,y1)

  13. 3x+4y-12=0 (4,0) 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 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)

  14. Solution Cont. r=  5 and cos =(-4/5) and sin =(3/5) Let the other end of diameter be (x1,y1) x1=4+5 (-4/5)=0 y1=0+ 5 (3/5)=3 for r=5 or x1=4-5 (-4/5) =8 y1= 0-5 (3/5) =-3 for r=-5 Other end of the diameter is either(0,3) or (8,-3)

  15. Solution Cont. 3x+4y-12=0 (0,3) (4,0) (x-0)(x-4)+(y-3)(y-0)=0 (8,-3) or x(x-4)+y(y-3)=0  x2+y2-4x-3y=0 (x-8)(x-4)+(y+3)(y-0)=0 x2+y2-12x+3y+32=0

  16. General Form Standard form: (x-xc)2+(y-yc)2=R2 ………….(1)  x2+y2-2xcx-2ycy+ xc2+ yc2- R2=0 In general formit is written as: x2+y2+2gx+2fy+ c=0 , where g, f, c R Comparing with (1) g=-xc, f=-yc c= xc2+ yc2- R2

  17. General Form x2+y2+2gx+2fy+ c=0 represents circle having center at (-g,-f) general form of circle and radius= 1. g2+f2-c>0 real circle (R>0) ? 2. g2+f2-c=0  Point circle (R=0) 3. g2+f2-c<0  Imaginary circle (R<0) ?

  18. Illustrative Problem Find the centre & radius of the circle 2x2+2y2-6x+6y-5=0 Centre (3,-3)? Solution: Note:- center is (-g,-f) when co-eff of x2 & y2 is 1 Make co-eff of x2 & y2 as 1 x2+y2-3x+3y-(5/2)=0 g=-(3/2); f=3/2;c=-5/2 Centre(3/2,-3/2)

  19. 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 x2+y2+2gx+2fy+c=0 be the equation of the circle. As it passes through (1,0)  2g+c=-1------(i) 12+02+2g.1+2f.0+c=0 As it passes through (2,-7) 22+(-7)2+2g.2+2f.(-7)+c=0  4g-14f+c=-53------(ii) As it passes through (8,1) 82+12+2g.8+2f.1+c=0  16g+2f+c=-65------(iii)

  20. Solution Cont. 2g+c=-1------(i) 4g-14f+c=-53------(ii) 16g+2f+c=-65------(iii) (ii)-(iii)  -12g-16f=12 (ii)-(i)  2g-14f=-52 or, 3g+4f = -3 or, g-7f = -26 By Solving  c=9 g=-5 ,f=3 Equation of the circle is: x2+y2-10x+6y+9=0 For (9,-6) : L.H.S= 92+(-6)2-10.9+6.(-6)+9 = 0 Four points are Concylic

  21. Condition For 2nd Degree Equn to Represent Circle General form of 2nd degree equn:- ax2+by2+2gx+2fy+2hxy+c=0---(1) where a,b,g,f,h,c  R It represents curves of different kind 2gx+2fy+c=0 AS : 1. for a=b=h=0 (and g ,f  0)  equn (1) be straight line

  22. Condition For 2nd Degree Equn to Represent Circle General 2nd degree equation: ax2+by2+2gx+2fy+2hxy+c=0---(i) General equation of the circle :x2+y2+2gx+2fy+c=0-------------(ii) So general 2nd degree equation will represent circle iff 1. Co-eff of x2=Co-eff y2 0  a = b  0 By Comparing (i) and (ii) 2. Co-eff of xy=0  h=0

  23. Illustrative Problem Given that ax2+by2+2gx+2fy+c=0 represents circle. Then find the radius of the circle? Solution: As ax2+by2+2gx+2fy+c=0---(1) represents a circle a=b  0 Equation (1) can be written as

  24. L2=0 L3=0 L1=0 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 L1 x+y-6=0 L2  2x+y-4=0 L3 x+2y-5=0 p3 L1L2+  L1L3+  L3L2=0 p1 What does it represent? p2

  25. Solution Cont. p3 (x3,y3) L3=0 L2=0 L1=0 p1 p2 (x1,y1) (x2,y2) f(x,y)=L1L2+  L1L3+  L3L2 (1) f(x,y)= (x+y-6)(2x+y-4)+(x+y-6)(x+2y-5) +(x+2y-5)(2x+y-4) x1+y1-6=0 and 2x1+y1-4=0 As P1(x1,y1) lies on L1=0 and L2=0 By putting P1(x1,y1) in (1)

  26. p3 (x3,y3) L3=0 L2=0 L1=0 p1 p2 (x1,y1) (x2,y2) Solution Cont. x1+y1-6=0 and 2x1+y1-4=0 f(x1,y1)= (x1+y1-6)(2x1+y1-4)+(x1+y1-6)(x1+2y1-5) +(x1+2y1-5)(2x1+y1-4) = 0x0+ x0x(x1+2y1-5) +(x1+2y1-5)x0 = 0  P1 lies on f(x,y) Similarly it can be proved that P2 and P3 also lies on f(x,y).

  27. p3 (x3,y3) L3=0 L2=0 L1=0 p1 p2 (x1,y1) (x2,y2) Solution Cont. • points of intersections of L1=0, L2=0 and L3=0 lies on curve f(x,y)=0 P1,P2,P3 lies on f(x,y)=0 Represents curve, which passes through the point of intersection of L1=0, L2=0 and L3=0 L1L2+  L1L3+  L3L2=0

  28. Solution Cont. L1L2+  L1L3+  L3L2=0 (x+y-6)(2x+y-4) +(x+y-6) (x+2y-5) +(x+2y-5)(2x+y-4)=0 1.Co-efficient of x2=co-efficient of y2 2++2=1+2+2-------(1) Will represent circle 2.Co-efficient of xy=0 3+3+5=0------------(2) From (1) and (2): =1 ,= -6/5

  29. Solution Cont. L1L2+  L1L3+  L3L2=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 • x2+y2 -17x-19y+50 = 0 is the equation of the circle

  30. P(x,y) R  C(xc,yc) (R sin) yc+Rsin w.r.t x-axis yc xc (R cos) xc+Rcos Parametric Form For the circle (x-xc)2+(y-yc)2=R2 point P(x,y) on circle represented as: x= xc + R cos  . y= yc + R sin  =y collectively represents all the points on circle for different values of : 0  360 =x

  31. y P(x,y) . R R sin  x R cos Parametric Form (x-xc)2+(y-yc)2= R2 Parametric form Special case: For the circle x2+y2=R2 Parametric form:-

  32. Illustrative Problem Find the parametric equation of the circle x2+y2-4x-2y+1=0 Solution: 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 )

  33. Class Exercise

  34. 2x-3y+12=0 Point of intersection of x+4y-5=0 and 2x-3y+12=0 Center Of Circle x+4y-5=0 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: 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

  35. Class Exercise - 2 • Find the parametric form of the circle: x2+y2+px+py=0. Solution: • x2+y2+px+py=0. Parametric Form of Equation

  36. 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 L1 x+y-2=0 L2 3x-4y-6=0 L3  x-y=0 Equation of circle passes through point of intersection of L1, L2 and L3 is

  37. Solution Cont. Equation (i) will represent circle if (a) co-efficient of x2 = co-efficient of y2 (b) co-efficient of xy = 0 from (2):  =-25/7

  38. 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  x2+y2–4x+14y–12=0

  39. C(0,6) B(4,0) O(0,0) 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: Diameter,asBOC=900 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, x2+y2–4x–6y=0

  40. S2=0 . S1=0 . C2(2,-5) C1(-3,7) Class Exercise - 5 • Find the equation of circle, the end-points of whose diameter are the centers of the circles x2+y2+6x– 14y–1=0andx2+y2–4x+10y–2=0. Solution: Diameter (x+3)(x–2)+(y–7)(y+5)=0 or, x2+y2+x–2y–41=0

  41. 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 • x2+y2+2gx+2fy+c=0 ...(i) -3g+4f+1=0 ...(ii) Centre : (–g,–f) 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

  42. Solution Cont. x2+y2+2gx+2fy+c=0…..(i) c=–3; f=–1; g=–1 • Equation of the circle is: • x2+y2–2x–2y–3=0

  43. B(2,2) r=1 A(1,1) Class Exercise - 7 Find the equation of the circle which passes through (1, 1), (2, 2) and radius is 1. Solution: . As it passes through P(1,1) C(h,k)  (h–1)2+(k–1)2=1 h2+k2–2h–2k+1=0 ...(i) As it passes through Q (2,2)  (h–2)2+(k–2)2=1 h2+k2–4h–4k+7=0 ...(ii) (i)—(ii) h+k=3  k=3–h

  44. Solution Cont. h2+k2–2h–2k+1=0 ...(i) h2+k2–4h–4k+7=0 ...(ii) k=3–h From (i): h2+(3–h)2–2h–2(3–h)+1=0 h=1,2  2h2–6h+4=0 h2–3h+2=0 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

  45. Class Exercise - 8 If one end of diameter of circle x2+y2–4x–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)

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

  47. x+y=2 2x+2y=3 Solution P(1,1) Observe: 1. x+y=2 and 2x+2y=3 are the parallel lines. Q 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.

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

  49. P(1,1) x+y=2 Q 2x+2y=3 Solution Cont. 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  4x2–7x+4y2–7y+6=0

  50. Class Exercise - 10 • If the equation of a circle is ax2+(2a–3)y2–4x–1=0 then the centre is • (2,0) (b)(2/3,0) (c) (-2/3,0) (d) None of these Solution: If ax2+(2a–3)y2–4x–1=0 represents circle  a=3 a=2a–3 • Equation of circle is: 3x2+3y2–4x–1=0  x2+y2–(4/3)x–1=0 Center: (2/3,0)

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