A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

ChillOut

  • one year ago

Re-asking a question that I made 2 years ago and couldn't get to an answer. Maybe now someone can reach it! Two circles, C and Cr intersect each other. C's radius is 1, while Cr's radius is "r". Their tangent lines form an internal angle of 120° at each intersection point. 1) Calculate the distance "d", which is the distance between the centers of both circles, in terms of r. 2) Find where d has a maximum or a minimum. 3) Find the area of the intersection in terms of π

  • This Question is Closed
  1. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1437616135891:dw|

  2. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1437616351195:dw|

  3. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I'm starting to think that this is unsolvable.

  4. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    All right, so I found this: http://www.had2know.com/academics/intersection-angle-two-circles.html Is this of any help?

  5. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I swear if I get an answer to this I'll become a math teacher.

  6. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Cutting out 30 degrees from either ends of the tangent lines, it is easy to see that the angle made by radii segments at the intersection point is 60 degrees : |dw:1437618536980:dw|

  7. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    How do I prove that it's 60°?

  8. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    If it's right, I can just go on with implicit differentiation, yeah?

  9. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Easy : draw a better diagram that actually represents 120 degrees between the tangents, then argue that the perpendicular to tangent line at the point of tangency passes through the center

  10. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Oh, yeah. I see it now. Now I think I can finish it.

  11. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I was failing to notice how the radii and the tangents connected to form a triangle.

  12. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    We don't really need calculus for second part, we may use the fact that \(d\) is positive and the max/min value of \(d\) occurs at the same point as \(d^2\)

  13. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    |dw:1437619117612:dw|

  14. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Yeah, I did the whole d'(r) and found 1/2 as well.

  15. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    *min value for \(d\) occurs at \(r = 1/2\)

  16. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I'm getting d^2 = 1 + r^2 + r

  17. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Just for future reference, leaving the derivative here: \(\frac{dd}{dr}=(2r-1)*(2(r²-r+1)^{0.5}\))

  18. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Law of Cosines c^2 = a^2 + b^2 - 2*a*b*cos(C) d^2 = 1^2 + r^2 - 2*1*r*cos(120) d^2 = 1^2 + r^2 - 2*1*r*(-0.5) d^2 = 1^2 + r^2 + r

  19. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @jim_thompson5910 , why is the angle between the radii 120º in your picture?

  20. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    @jim_thompson5910 looks 120 degrees has to be the other angle : |dw:1437619755534:dw|

  21. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Now I am confused, If they are orthogonal to the tangent lines, they should have the same angle between them, right?

  22. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    look at the attached pic I'm posting angle CAF = 90 degrees since the circle is tangent at point A angle GAD = 90 degrees (same reason) angle DAF = 30 degrees because it's the solution so 90+x = 120 angle DAC = (angle DAF) + (angle CAF) angle DAC = (30) + (90) angle DAC = 120

  23. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1437619928573:dw|

  24. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    again, the construction is incorrect as 120 degrees has to be the internal angle as per the question

  25. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1437619979044:dw|

  26. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh did I flip the angles?

  27. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    I think so, after flipping the angles, ur construction also gives the same expression \(d^2 = 1+r^2-r\)

  28. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1437620217558:dw| This one is supposed to have 120°

  29. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    So, by using the law of cosines and assuming it's 60° it gives @ganeshie8's expression

  30. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    It is easy to prove using that new better picture, lets do it

  31. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    |dw:1437620436117:dw|

  32. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    |dw:1437620527012:dw|

  33. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Angle between radius and tangent = 90 therefore angle between radius and other circle's tangent = 120-90 = 30 similar reasoning yields the other angle also 30

  34. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    |dw:1437620634836:dw|

  35. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    so the angle between radii segments = 120-30-30 = 60

  36. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Yes, pretty nice! We already have the critical point for the radii. Now I need to remember how to answer c) without resorting to calculus

  37. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Just find the area of both sectors segments and add them?

  38. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    calculus is always there if we forget geometry :) lets see if this can be done smartly w/o calc

  39. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    1 Attachment
  40. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    1 Attachment
  41. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    thing is... I need to find Beta.

    1 Attachment
  42. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Or do I?

  43. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Well, I remember the area of a circle's sector. It's given by \(A=r²*arccos(d/2r)\) Where d is the distance between the centers and r the radius of the circle in question.

  44. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    you want to find the area for general case, in terms of \(r\), right ?

  45. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    This should be ideal.

  46. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Well, I'm a lackluster in geometry, as you can see.

  47. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    your method works, start by finding this central angle, \(x\) : |dw:1437623030778:dw|

  48. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    once you have \(x\), you can find the area of sector and subtract the area of triangle from it to get : |dw:1437623216203:dw|

  49. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    you can do the same for the other piece also! looks good!

  50. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    The line that cuts through both intersections also divides the angles equally, right?

  51. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Nope.

  52. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    use base angles of isosceles triangle theorem

  53. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    |dw:1437623442647:dw|

  54. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    applying cosine law in that triangle with sides \(1, r, d\) : \[r^2 = 1^2+d^2-2d\cos(x/2)\] you can find \(x\)

  55. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Wouldn't I be able to solve for x only i I had a particular solution for r, which isn't the case?

  56. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    you just want to find the overlapping area for \(r=1/2\) is it ?

  57. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Yeah. I don't see any way to find a general solution as d is defined by r.

  58. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    applying cosine law in that triangle with sides \(1, r, d\) : \[r^2 = 1^2+d^2-2d\cos(x/2)\] \[r^2 = 1+(1+r^2-r)-2d\cos(x/2) \implies x = 2\arccos\left(\dfrac{2-r}{2d}\right)\]

  59. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    \(d\) is just a number as you can see, we can always replace \(d^2\) by \(1+r^2-r\)

  60. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    All right, I need to calculate the sector's area.

  61. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \((x/2\pi)*\pi*r²\)

  62. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Area of shaded region in circle of radius \(1\) : \[\pi*\dfrac{x}{2\pi} = \arccos\left(\dfrac{2-r}{2d}\right)\]

  63. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Just subtract the triangle area and we have the segment, right?

  64. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Yes, just algebra..

  65. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    So, \(A_{segment}=arccos(\frac{2-r}{2d})-\frac{1*1*sin(x)}{2}\)

  66. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    which in turn is \(arccos(\frac{2-r}{2d})-sin(2arccos(\frac{2-r}{2d})) \)

  67. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Oops! Missed /2 in the second part.

  68. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Looks good to me! Area of shaded region in circle of radius \(1\) : \(\large \arccos(\dfrac{2-r}{2d})-\dfrac{\sin(2\arccos(\frac{2-r}{2d}))}{2} \) |dw:1437625381250:dw|

  69. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Now for the other sector... Which will be in terms of "r"

  70. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    |dw:1437625666331:dw|

  71. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    do we get \[y=2\arccos\left(\dfrac{2r^2-r }{ 2dr}\right)\] ?

  72. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Yes.

  73. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Wait... Let me take a second look.

  74. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Yeah, that's what I got.

  75. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \(A_{segment}=r^{2}arccos(\frac{2r^{2}-r}{2dr})-\frac{r^{2}sin(2arccos(\frac{2r^{2}-r}{2dr}))}{2}\)

  76. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    For the circle of radius r. Sorry for the small TeX text

  77. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Then, adding the segments' areas, we have \(arccos(\frac{2-r}{2d})-\frac{sin(arccos(\frac{2-r}{2d})}{2}+r²arccos(\frac{2r²-r}{2dr})-\frac{r²sin(2arccos(\frac{2r²-r}{2dr}))}{2}\)

  78. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I guess this is it!

  79. ChillOut
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    P.S: Missed a "2" before the first arc cosine.

  80. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Looks neat!

  81. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.