Part r A A A 1 Mark Part r B B B 2 Marks Mark P t ar t t C C 5 Mar M ks Part r E 4 Marks Mark Tot To a t l

Transcription

1 Part Part P t Part Part Total A B C E 1 Mark 2 Marks 5 Marks M k 4 Marks CIRCLES 12 Marks approximately

2 Definition ; A circle is defined as the locus of a point which moves such that its distance from a fixed point is constant.

3 A Second degree equation in x and y ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 represents a circle if a = b and h = 0 Eg : Find the values of p and q If px 2 + 2y 2 + (q 1) 1)xy 4x + 2y 1= 0 represents a circle. p = 2, q 1 = 0, q = 1

4 Equation of the circle with centre (h, k) and radius r is ( x h ) 2 + ( y k ) 2 = r 2 c r (h, k) If the centre is origin, o r (o, o) x 2 + y 2 = a 2 (o, o)

5 Eg : Find the equation of the circle with radius 5 units whose two diameters are the lines 2x + 3y = 5 and 7x 2y = 5

6 Solving the diameters we get. x = 1, y = 1 given r = 5.. The equation of the circle will be ( x 1 ) 2 + ( y 1 ) 2 = 5 2 x 2 + y 2 2x 2y 23 =0 (1, 1) C

7 The parametric equations of the circle x 2 + y 2 = r 2 are x = rcosθ, y = rsinθ for (x h) 2 +(y k) 2 =r 2 are x = h + rcosθ, y = k + rsinθ Eg ; Find the Area of the Circle X = cosθ, y = 3 + 2sinθ Area = 4 4π π sq. units

8 General equation of a circle X 2 + y 2 + 2gx + 2fy + c = 0 Centre = ( g, f) And radius = g 2 + f 2 c Points to be remembered If a Circle passes through h (0,0) then c = 0 If the centre lies on x axis, then f = 0

9 If the centre lies on y axis then g = 0 If the circle touches x axis then g 2 = c If the circle touches y axis then f 2 =c If a circle touches both the axes then g 2 = f 2 = c (x a) 2 + (y a) 2 = a 2

10 2. Find the equation of the circle passing through the origin, having its centre on x axis with radius 2 units. C= 0, f = 0, r = 2 => g 2 + f 2 c = 2 => g 2 = 2 g = ±2... Equation of the circle is X 2 + y 2 ± 4x = 0

11 Conditions for 2 Circles to touch each other. Externally Internally P r 1 r2 A B A B P AB = r 1 + r 2 AB = r 1 r 2

12 Eg. Consider the circles, X 2 + y 2 2x 6y 134 = 0 and X 2 + y 2 14x + 10y + 70 = 0 A = (1,3), B = (7, 5) r 1 = r 2 = 2 = 12 AB = AB = 10 = r 1 r 2... They touch internally

13 Equation of the Circle with (x 1, y 1 ) & (x 2, y 2 ) as the ends of a diameter is (x x 1 ) (x x 2 ) + (y y 1 ) (y y 2 ) = 0 P (x 2, y 2 ) C (x 1, y 1 )

14 Eg :The equation of the circle passing through origin,,(, (a, 0) and (0, b) is (x 0) (x a) + (y 0)(y b) = 0 X 2 + y 2 ax by = 0 (0, b) 0 (a, 0)

15 Length of the chord of the circle Intercepted by x axis is 2 g 2 c Intercepted by y axis is 2 f 2 c Intercepted by a line ax + by + c = 0 is 2 r 2 p 2 p r

16 Eg. Find the length of the chord of the circle x 2 + y 2 6x 2y + 5 = 0 Intercepted by the line x y+1 = 0. Centre = (3, 1), r = 5 p = = Length of the chord = 2 r 2 - p 2 = 2 2 units

17 Results. 1.The equation of the tangent to the circle x 2 + y 2 + 2gx + 2fy + c = 0 at the point (x, y 1 ) on it is xx 1 + yy 1 + g(x+x 1 ) + f(y+y 1 ) + c = 0 P(x, y 1 ) C. ( g, f)

18 2. The condition for a line y= mx + c to be a tangent to the circle x 2 + y 2 = a 2 is c 2 = a 2 (1 + m 2 ) Point of Contact = a 2 m, a 2 P c c y = (mx+c +c) (0,0) a

19 Ex.1 Find K if the line 3x+2y = k is a tangent to the circle x 2 +y 2 = 10. 2y = 3x + k y = 3x + k 2 2 m = 3 3, c = k, a 2 = c 2 = a 2 (1+m 2 ) k= ± 130

20 2. Find the equation of the tangent tt to the circle x 2 +y 2 2x 4y 4 = 0 which are perpendicular to the line 6x 8y +1 = 0 Ltth Let the tangent tb be, 8x 6y+ k = 0 Or 8x + 6y k= 0 Radius = r distance Solving we get k = 10 or Required tangents are 4x + 3y +5 = 0 and 4x + 3y 25 = 0

21 Result Length of the tangent from a point P(x 1 y 1 ) to the circle x 2 + y 2 + 2gx + 2fy + c = 0 is PT = x 12 + y 12 +2gx 1 +2fy 1 +c T Power of a point P w.r.t a circle The Power of a Point P(x 1, y 1 ) w.r.t a circle is defined as Pc 2 r 2 = x y gx 1 + 2fy 1 + c c P(x 1, y 1 )

22 If the power of the point > 0 the Point lies outside the circle < 0 the Point lies inside the circle = 0 the Point lies on the circle Eg. Power of the point (2, 7) w.r.t the circle X 2 +y 2 + 2x 6y + 2 = 0 is x2 6x7 + 2 = 17>0... (2, 7) lies outside the circle.

23 Radical Axis Radical axis of two circles is the locus of a point which moves such that its powers w.r.t two given circles are equal. The radical axes of two circles s 1 = 0 and s 2 = 0 is given by 2 (g 1 g 2 ) x + 2 (f 1 f 2 ) y = c 1 c 2 i.e: S 1 S 2 = 0

24 Result: The Radical axis of two circles is perpendicular to the line joining the centres. R.A Slope of R.A.x slope of c 1 c 2 = 1

25 Rdi Radical axes of 3 circles are concurrent if the centres are not collinear S 1 = 0 S 2 = 0 Radical Center S 3 = 0

26 R.A is the common tangent To find the length of the common chord = 2 r 2 p 2 p L RA is the common Chord

27 Orthogonal Circles Condition for two circles x 2 +y 2 +2g 1 x+ 2fy 1 +c 1 = 0and x 2 +y 2 + 2g 2 x+ 2fy 2 +c 2 = 0 to cut h ll i 2g 1 g 2 + 2f 1 f 2 = c 1 + c 2 orthogonally is 2g P A

28 Ex. If the Circles x 2 +y 2-2x 2x- 2y -7 =0 and x 2 +y 2 +4x +Ky +1=0 are orthogonal find K. 2g 1 g 2 + 2f 1 f 2 = c 1 + c 2 2(-1) )( (+2) + 2 (-1) )( K ) = Simplifying, we get K = 2

29 Example: 1. Find the equation of the circle passing through the point (2, 1) and having its centre on the line 2x+ y =1 and cutting the circle x 2 +y 2 +2x +6y+5= 0 orthogonally Let the equation of the circle be x 2 +y 2 +2gx+ 2fy +c =0 As (2, 1) lies on it, 4g 2f+c = 5 5 (1)

30 Thecentre ( g, f) lies on 2x+y =1... 2g f = 1.(2) If cuts x 2 +y 2 +2x +6y+5 orthogonally 2(g) (1) + 2f(3) = c+5 2g+6f 2g+6f c = 5.(3) Solving, we get, g = 2, f=3 and c = 9... Required circle is x 2 +y 2-4x +6y+9 = 0

31 2. Find the equation of the circle which cuts the circles x 2 +y 2 8x+ 3y +21 =0 x 2 +y 2 7x+ 7 5y +28 =0 x 2 +y 2 +x + y 16 =0 orthogonally. S 1 = 0 S 2 = 0 S 3 = 0

32 Sincetherequired circle cutsgiven3 circles orthogonally, we have 8g+3f+21 = 0 7g+5f+28 = 0 g + f 16 = 0 solving these g = 3,, f = 5 and c = 18 Required circle is x 2 +y 2 6x 10y+18 = 0

33 Inverse Trigonomeric Functions Part A 1 Marks Part B 2 Marks Part C 3 Marks

34 Before we go to the Inverse Trigonometric functions, we shall know about the inverse of a function. The Function f: A B has inverse only if f is both one one and onto (bijective) ie every different elements of A have different images and all theelements elements of B must be the image of atleast one element ofa

35 Question is whether the inverse exist for trigonometric functions. Answer is No.

36 As we know from the diagram, for 2 different values of θ, the values of sinθ, cosθ, and tanθ, are same... Trigonometric functions are not one one To make trigonometric functions one one, we restrict the domain θ so that all the trigonometric functions becomes 1 1 and onto.

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38 From the above diagram, we see that Sinθ θ is 1-1 if θ lies in the interval - 90 to 90 Similarly cosθ θ is 1-1 if θ lies in the interval 0 to 180 tanθ θ is 1-1 if θ lies between and 90

39 Inverse sine function Definition: If x= sinθ and θ 2 2 then θ = sin 1 x, where 1 x 1 is called inverse sine function Inverse cosine function Definition: If x= cosθ and 0 θ then θ = cos 1 1 x, where 1 1 x 1is called inverse cosine function

40 Inverse tangent function Definition: If x= tanθ and < θ < 2 2 then θ = tan 1 x, where < x < is called inverse tangent function Inverse cotangent function Definition: If x= cotθ and 0 < θ < then θ = cot 1 x, where < x < is called inverse cotangent function

41 Properties of Inverse Trigonometric functions. I 1.sin 1 ( x) = sin 1 x cosec 1 ( x) = cosec 1 x 2. cos 1 ( x) = π cos 1 x sec 1 ( x) = π sec 1 x 3. tan 1 ( x) = tan 1 x cot 1 ( x) = π cot 1 x

42 II 1. cosec 1 (x) = sin 1 Or sin 1 x= cosec 1 x 2. sec 1 x = cos 1 cos 1 x= sec 1

43 3. cot 1 x = tan 1 Or tan 1 x = cot 1

44 III 1. sin 1 x + cos 1 x = π 2 2. tan 1 x + cot 1 x = π 2 3. sec 1 x + cosec 1 x = π 2

45 IV 1. sin 1 (sin sinθ) = θ only if θ cos 1 (cos cosθ) = θ only if 0 θ 3. tan 1 (tan tanθ) = θ only if < θ < cot 1 (cot cotθ) = θ only if 0 < θ <

46 V 1. tan 1 x +tan 1 y = tan 1 x+ y x 0, y 0 and xy<1 2. tan 1 x + tan 1 y = π + tan 1 1 x + y x>0, y>0 and xy>1

47 3. 2 tan 1 x =tan 1 2x 4. sin 1 x +sin 1 y = sin 1 (x 1 y 2 + y 1 x 2 ) 5. cos 1 x +cos 1 y = cos 1 (xy 1 y 2 1 x 2 )

48 1. Evaluate sec 1 ( 2 ) Solution: sec 1 ( 2 ) = π - sec 1 2 = π - cos 1 (1/ 2) = π π/4 = 3π/4

49 2.Evaluate: cosec 1 (2) Solution: cosec 1 (2) = sin 1 (1/2) = π/6 3. Evaluate: cot 1 ( 3) Solution: cot 1 ( 3) = π cot 1 ( 3) = π tan 1 (1/ 3) = π π/6 = 5π/6

50 4. Find the value of sin[π/2 sin -1 (- 3/2)] Solution: sin[π/2 sin -1 (- 3/2)] = sin[90 ( sin -1 ( 3/2)] =sin[90+60] = cos60 = 1/2

51 5.Evaluate: sin[cos -1 (3/5)] Solution: Let cos -1 (3/5) / ) = θ... Cosθ = 3 = adj 5 hyp Note: In any right angled triangle.

52 Hyp 2 = opp 2 + adj 2 opp 2 = hyp 2 adj 2 adj 2 = hyp 2 opp 2... Opp 2 = hyp 2 adj 2 = = 16 Opp = 4, Hence sinθ = opp = 4 hyp 5

53 6. Find the value of cos 1 (cos585) Solution: cos 1 (cos585) = cos 1 (cos( )) = cos 1 (cos225) = cos 1 (cos( )) = cos 1 ( cos45) = π - cos 1 (cos45) = π π/4 = 3π/4

54 7.Evaluate cos(2cos 1 4/5) Solution: Let cos 1 4/5 = θ => cosθ = 4/5 We find cos2θ Now cos2θ = 2cos 2 θ 1 = 2(4/5) 2 1 = 2(16/25) 1 = (32/25) 1 = (32 25)/25 = 7/25 Evaluate: cos(2sin sin 1 4/5) Evaluate: cos(2tan 1 4/3)

55 8. Evaluate: cos(1/2 sin 1 5/13) Solution: Let sin 1 5/13 = θ sinθ = opp = 5 hyp 13 adj 2 = hyp 2 opp 2 = = Adj = 12 => Cos θ = 12/13 Now, we find cosθ/2 Cosθ/2 = 1+cos 1+cosθ/2... Cosθ/2 = 5/ 26

56 9.Prove that 2tan 1 1/3 + tan 1 1/7 = π/4 Solution: Using 2tan 1 x = tan 1 (2x/1 x x 2 ) 2tan 1 1/3 = tan 1 2(1/3) 1 (1/3) 2 = tan 1 2/3 1 (1/9) = tan 1 2/3 8/9 = tan 1 ( 2/3 x 9/8) = tan 1 3/4

57 Now tan 1 3/4 + tan 1 1/7 =tan 1 3/4 + 1/7 1 (3/4)(1/7) =tan 1 (21+4)/28 = tan 1 25/28 1 3/28 25/28 = tan 1 1 = π/4 Prove that 2tan 1 1/3 + tan 1 4/3 = π/2 Prove that2tan 1 1/3 + sin 1 1/5 2 = π/4

58 10. If tan 1 x+ tan 1 y= π/4 Prove that X+y+xy = 1 Solution: tan 1 + tan 1 y= π/4 tan tan 1 x + y = π/4 x+y/1 xy xy = tan π/4 =1 X+y=1 =1 xy => x+y+xy = 1

59 11. tan 1 x+ tan 1 y+tan 1 z= π Prove that X+y+z = xyz Solution: tan 1 x+ tan 1 y= π tan 1 z tan tan 1 x+ y = π tan 1 z x+y/1 xy xy = tan (180 tan 1 z) = tan( tan(tan 1 z) = z x+y= z(1 (1 xy) = z+ + xyz... x+y+z = xyz If tan 1 x+ tan 1 y+tan 1 z= π/2 Prove that xy+yz+zx = 1

60 12. If sin 1 x+ sin 1 y = π/2 /, Show that x 2 + y 2 =1 Solution: sin 1 x= π/2 / sin 1 y sin 1 x= cos 1 y sin 1 x= sin 1 1-y y 2 x = 1-y 2 => x 2 = 1 y 2. x 2 + y 2 =1.. y

61 13. If sin 1 x+ sin 1 y+sin 1 z= π/2, Show that x 2 + y 2 + z 2 + 2xyz=1 Solution: sin 1 x+ sin 1 y= π/2 sin 1 z sin 1 x + sin 1 y = cos 1 z Let sin 1 x= A => x = sina, cosa = 1 Let sin 1 y = B => y = sinb, cosb = 1 Now, A+B = cos 1 z cos ( A+B) = cos(cos 1 z ) cosacosb sinasinb = z 1-x 2 1-y y 2

62 ( 1-x 2 )( 1 y 2 ) xy = z ( 1-x 2 )( 1 y y 2 ) = xy + z Squaring both sides, (1 x 2 )(1 y 2 ) = (xy+z) 2 1 y 2 x 2 + x 2 y 2 = x 2 y 2 + z 2 + 2xyz... x 2 + y 2 + z 2 + 2xyz=1

63 14. If cos 1 x+ cos 1 y+cos 1 z= π, Show that x 2 + y 2 + z 2 + 2xyz=1 Solution: cos 1 x+ cos 1 y= π cos 1 z Let cos 1 x = A => x = cosa, sina = 1 Let cos 1 y= B => y = cosb, sinb = 1 Now, A+B = π - cos 1 z cos ( A+B) = cos(180 cos 1 z ) cosacosb sinasinb = cos ( cos 1 z ) 1-x x 2 1-y 2

64 xy ( 1 ( 1-x 2 )( 1 y 2 )= z xy + z =( 1 ( 1-x 2 )( 1 y y 2 ) Squaring both sides, (xy+z) 2 = (1 x 2 )(1 y 2 ) x 2 y 2 + z 2 + 2xyz =1 y 2 x 2 + x 2 y 2... x 2 + y 2 + z 2 + 2xyz=1

65 15. Solve for x sin 1 x+ sin 1 2x = 2π/3, Solution: Let sin 1 x= A => x = sina, cosa= 1 sin 2 A = 1 x x 2 Let sin 1 2x = B => 2x = sinb, cosb= 1 sin 2 B = 1 (2x) 2 = 1 4x 2 Given, A+B = 120 Cos (A+B) = cos120 cosacosb sinasinb = cos ( )

66 )( 1 4x 2 ) x(2x) = cos60 ( 1-x 2 )( 1 4x 2 ) = (2x 2 1/2 ) Squaring both sides, (1-x 2 )(1 4x 2 ) = (2x 2 1/2 ) 2 1 4x 2 x 2 + 4x 4 = 4x 4 +1/4 2(2x 2 )(1/2) 1 5x 2 = 1/4 2x 2 1 1/4 = 5x 2 2x 2 3x 2 = 3/4 x 2 = 1/4 ( 1-x 2 )(

67 16. Solve for x sin 1 x+ cos 1 2x = π/6, Solution: Let sin 1 x= A => x = sina, cosa= 1 sin 2 A = 1 x 2 Let cos 1 2x = B => 2x = cosb, sinb= 1 cos 2 B = 1 (2x) 2 = 1 4x 2 Given, A+B = 30 sin (A+B) = sin30 sinacosb + cosasinb = sin 30 sinacosb + cosasinb = 1/2

68 (x)(2x)+ ( 1-x 2 )( 1 4x 2 ) = 1/2 ( 1-x 2 )( 1 4x 2 ) = ( 1/2 2x 2 ) Squaring both sides, (1-x 2 )(1 4x 2 ) = (1/2 2x 2 ) 2 1 4x 2 x 2 + 4x 4 = 4x 4 +1/4 2(2x 2 )(1/2) 1 5x 2 = 1/4 2x 2 1 1/4 = 5x 2 2x 2 3x 2 = 3/4 x 2 = 1/4

69 17. Solve for x, cos 1 x sin 1 x= cos 1 x 3, Solution: Let cos 1 x= A => x = cosa, sina= 1 cos 2 A = 1 x 2 Let sin 1 x= B => x = sinb, cosb= 1 sin 2 B = 1 (x) 2 = 1 x 2 Given, A B = cos 1 x 3 cos(a B) = x x 3 cosacosb + sinasinb = x 3 x 1 x x 2 (x) = x 3

70 2x 1 x 2 x 3 = 0 x(2 1 x 2 3) = 0... X=0 or 2 1 x 2 3 = x 2 = 3 Squaring both sides 4(1 x 2 )= 3 1 x 2 = 3/4 => x 2 = 1 3/4 = 1/4... X=0 and x= ±1/2

71 18. Solve for x, tan 1 x+ 2cot 1 x = 2π/3 Solution: tan 1 x+ 2cot 1 x = 120 tan 1 x+ cot 1 x + cot 1 x = cot 1 x = 120 cot 1 x = = X = cot 30 = 3