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Circle in parabola

  1. Aug 6, 2006 #1
    Hi I am stuck on the following:

    Find the centre of a circle with a radius of 1 inscribed in the parabola [tex]y=x^2[/tex].

    I am kinda stuck:grumpy:

    http://img115.imageshack.us/my.php?image=csacacfu8.png
    the white line is length=1
     
    Last edited: Aug 6, 2006
  2. jcsd
  3. Aug 6, 2006 #2

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    I'll do a much simpler example and see if it helps you. Imagine you want to find the point where a unit circle centered at some point on the y-axis is tangent to the x-axis. Clearly we want the center at (0,1) or (0,-1). But another way to find this is to look at where a unit circle centered at (0,h) intersects the line y=0. This is found as the set of simultaneous solutions to:

    [tex](y-h)^2+x^2=1[/tex]
    [tex]y=0[/tex]

    which is just the set of solutions to:

    [tex] x^2+h^2=0[/tex]

    If |h|>1, there are no solutions (which reflects the fact that the circle doesn't touch the x-axis), if |h|<1, there are two solutions (the circle straddles rhe x-axis and intersects it in two points), and if h=1 or h=-1, there is one solution, where the circle is tangent, which is just what we expected. Can you figure out how to extend this to your question?
     
  4. Aug 6, 2006 #3
    errr..im still confused, although I do understand what youve done I dont know how to apply it.
     
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