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Finding intersections of tangents on circles

  1. Jan 16, 2008 #1
    1. The problem statement, all variables and given/known data

    A circle, with a center at the origin, and a radius of 2 has at least one tangent
    with a slope of root3/3 and at least one tangent with a slope of -root3/3 .
    Calculate all intersection or intersections of these tangents


    2. Relevant equations



    3. The attempt at a solution

    how do i even do this?!?!?
     
  2. jcsd
  3. Jan 16, 2008 #2
    What is the equation of your circle?
     
  4. Jan 16, 2008 #3
    i don't have one .....that's the problem.... that is the exact question word for word ....
     
  5. Jan 16, 2008 #4
    Yes you do, think back to how you can write the equation of the circle by just knowing it's position is at the Origin and that it's radius is 2.
     
  6. Jan 16, 2008 #5
    (0 - h)^2 + (0 - k)^2 = 4 ??
     
  7. Jan 16, 2008 #6
    Good try, but h & k = 0, not x & y.

    [tex]x^2+y^2=4[/tex]

    So, take the derivative ... and solve for y'.
     
  8. Jan 16, 2008 #7
    alright so 2x + 2y = 0
    and 2y = -2x
    then y = -2x / 2
    y = -x ........... so wtf does that have to do with finding the intersect
     
  9. Jan 16, 2008 #8
    You didn't even do it right so don't 'wtf', do it again and get y' if you still want help.

    "Implicit differentiation"
     
  10. Jan 16, 2008 #9
    AHHHHHHHHHH alrighty 1 sec lemme calculate this
     
  11. Jan 16, 2008 #10
    okay

    i got dy/dx = -2x/2y ......... then what??
     
  12. Jan 16, 2008 #11
    [tex]y'=\frac{-x}{y}[/tex]

    We're told one of our slopes is

    Tangent 1 = [tex]\frac {1}{\sqrt{3}} =\frac {-x}{y}[/tex] and also gives us our coordinate [tex](-1,\sqrt{3})[/tex]

    Now just use the Point-slope equation for this one and the other one whose slope can obviously be found the same way.

    And to find where the slopes intersect, set them equal to each other and solve for the x position.
     
    Last edited: Jan 16, 2008
  13. Jan 16, 2008 #12
    um ...... i thought the tangents slopes were root3 / 3 and negative root3 / 3
     
  14. Jan 16, 2008 #13
    so then would root3/3 = -x/y ???
    and -root3/3 = -x/y??
     
  15. Jan 16, 2008 #14
    Your slopes are rationalized while the ones I used are not. It's easier to type into the calculator or to write.
     
  16. Jan 16, 2008 #15
    Tell me what your coordinates would be if you were to use those values?

    If you plug in those values into your original equation, will it equal 4? Rationalizing isn't always a good thing to do.
     
  17. Jan 16, 2008 #16
    ahhhhhh ic ........... so root3/3 = 1/root3
     
  18. Jan 16, 2008 #17
    Yep, and it's definitely best to use the unrationalized version so that you can also determine what your x-points will at those tangents.
     
  19. Jan 16, 2008 #18
    y – y1 = m(x – x1) point slope formula right???
    so i have the point (-1,root3) from the first slope
    now i find the point from the second slope right??
    and it should be ( 1,root3) >??
     
  20. Jan 17, 2008 #19
    Yes, and since this a circle, only 1 point will change.
     
  21. Jan 17, 2008 #20
    alright so i plug in those co-ordinates into the point slope equation?? and that should find me my intersect point right?
     
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