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Circle - Line Intersection

  1. Dec 3, 2006 #1
    1. The problem statement, all variables and given/known data

    I can't find the intersection for Line y = 3/4x - 35/4
    and Circle y^2 + x^2 = 25


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 3, 2006 #2
    set them equal to each other.

    [tex] y^{2} = 25 - x^{2} [/tex]


    [tex] y =\pm \sqrt{25-x^{2}} [/tex]
     
    Last edited: Dec 3, 2006
  4. Dec 3, 2006 #3
    how do i get rid of the square on y after i do that
     
  5. Dec 3, 2006 #4
    oh sry didnt see that
     
  6. Dec 3, 2006 #5
    ah can someone do this for me please, i'm stuck
     
  7. Dec 3, 2006 #6

    radou

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    Homework Helper

    Courtrigrad pretty much solved it for you already - all you have to do is set the equations equal, since the equation of the line is already given in the form y(x) = ...

    Edit: draw a sketch first, this will pretty much solve your problem. :wink:
     
    Last edited: Dec 3, 2006
  8. Dec 3, 2006 #7
    i kno but i gor to 0 = -3/4x + 5ix + 35/4 and i don't really kno how to solve that
     
  9. Dec 3, 2006 #8

    radou

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    OK, let's slow down. Do you know how to make a sketch of the circle y^2 + x^2 = 25, and the line y = 3/4x - 35/4 ? The circle is centered at the origin, with radius 5, and you can sketch down the line easily by finding the points of intersection with the x and y axis (i.e. setting y = 0, and x = 0). What does that sketch tell you?
     
  10. Dec 3, 2006 #9
    well i graphed it and it showed no intersection
     
  11. Dec 3, 2006 #10
    now i just need to show that the circle and the line don't intersect algrebraecally or however you spell it.
     
  12. Dec 3, 2006 #11

    radou

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    Yes, after reading the posts above again, you'll end up with a quadratic equation which has no real solutions, which is what you need to show.
     
  13. Dec 3, 2006 #12
    so wait this function 0 = -3/4x + 5ix + 35/4 has no real solutions because there is a complex number right?
     
  14. Dec 3, 2006 #13

    Hurkyl

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    Gold Member

    Incorrect -- sometimes complex equations can have real solutions. You have to show the solutions are not real. (say, by solving it)
     
    Last edited: Dec 3, 2006
  15. Dec 3, 2006 #14
    I'm wondering how you arrived at that equation, because I don't think you should end up with a quadratic equation with complex coefficients in this problem.
     
  16. Dec 3, 2006 #15
    well if u set the y values to equal each other
    3/4x + 35/4 = sqr root of (25 - x^2)

    you get 0 = -3/4x + 5ix + 35/4
     
  17. Dec 3, 2006 #16
    Ummm... I don't think thats what you get. I don't see where the 5ix comes from at all.
     
  18. Dec 3, 2006 #17
    well u change it into sqr(25) x sqr(-x^2)
    then u sqroot the 25 to 5 and the sqr root of -x^2 is sqr(-1) x sqr(x^2)
    so its 5ix
    i = sqr(-1)
     
  19. Dec 3, 2006 #18
    You can't do that.
     
  20. Dec 3, 2006 #19
    the square root of -1 is called i
    its always called i
     
  21. Dec 3, 2006 #20
    Im only 14......
     
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