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Hey i thought that this is tricky but who knows

  1. Nov 21, 2007 #1
    [SOLVED] hey i thought that this is tricky but who knows

    hey guys iam trying to solve this question
    and this what i do have
    https://www.physicsforums.com/attachment.php?attachmentid=11650&stc=1&d=1195680084

    *i know that the radius is 1
    *then the x = 0
    *i think this question can be solved in two equations like those which are applied in knowing the center of mass ( in physics )
    so guys can you help me in this question
     

    Attached Files:

  2. jcsd
  3. Nov 21, 2007 #2

    berkeman

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    Staff: Mentor

    Thread moved to homework help.

    What can you show us about the equation for a circle, and how can you relate that to the equation shown for the parabola?
     
  4. Nov 22, 2007 #3
    hey why did you move that.,please replace it as its first condition, because first it is not a home work, second the circle equation is
    ((x^2)-x)+((y^2) - y)= R^2
     
  5. Nov 22, 2007 #4
    and why did i take a warning if you don't want to help just ignore, not to give me a warning
     
  6. Nov 22, 2007 #5

    quasar987

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    Science Advisor
    Homework Helper
    Gold Member

    You get warnings from posting homework TYPE problems in the wrong section.

    It's the mentor's job to give those warning it's not personal. It's like telling a police officier "hey, if you don't approve of *insert illegal activity of preference* just ignore, not to give me a warning."
     
  7. Nov 23, 2007 #6

    berkeman

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    Staff: Mentor

    Quasar explained why the thread was moved to Homework Help (thanks).

    Now that you have the equation for the circle, since the circle touches the parabola at just two places as shown, how can you set up the simultaneous equations to give you the x,y coordinates where they touch? Remember, because of the symmetry of the situation, you only really have to solve for one of the two touching points in order to get them both....
     
  8. Nov 24, 2007 #7

    Gib Z

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    Homework Helper

    And the equation of that circle is actually [tex]x^2 + (y- y_1)^2 =1 [/tex] where (0, y_1) is the center.
     
    Last edited: Nov 24, 2007
  9. Nov 29, 2007 #8
    thanks for interesiting i solved it and the answer is (0 , 1.25). first i solved for the equation of a circle and then i got another equation which is ( the tangent * thenormal=-1)
    whith a lot of my friend's help.
    thanks again
     
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