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A circle of radius R is centered at the origin.

  1. May 22, 2005 #1
    A circle of radius R is centered at the origin. The y-intercept (positive) is at (0, R). As we shift the circle to the left (or right) by n on the x-axis, the y-intercept decreases. Are we able to determine the radius of the circle given a certain proportion between the x-shift and the decrease in y-intercept? Say, if x is shifted by 45, then the y-intercept decreases by 5, what is the radius?
     
  2. jcsd
  3. May 22, 2005 #2

    honestrosewater

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    I don't know what to do about the degenerative case, but could you solve x2 + y2 = (x + n)2 + (y + m)2 given only n and m?
    Edit: Maybe I do know what to do in the degenerative case- x = 0. So your example is y2 = 452 + (y - 5)2
     
    Last edited: May 22, 2005
  4. May 22, 2005 #3
    Intuitively, I'm pretty certain that if two circle's radii are different, then the decrease in y-intercept can't be the same, given the same displacement in X.
     
  5. May 22, 2005 #4
  6. May 22, 2005 #5

    honestrosewater

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    I don't how many solutions there are for y2 = 452 + (y - 5)2. I was just considering that the circle's center C, the graph's origin O, and the y-intercept Y form a right triangle with line CY as the hypotenuse & radius, and the radius is constant.
    Edit: But consider that eventually CY = CO...
     
    Last edited: May 22, 2005
  7. May 22, 2005 #6
    Yes, I've already worked out (y - 5)^2 + 45^2 = y^2, but I wasn't sure, because a friend did it by drawing out things to scale, and got a different answer. I was afraid that my algebra or the whole concept was wrong.
     
  8. May 22, 2005 #7
    I'm tired as hell (3am here), so I may have misread the original problem, but I think a simpler formula for the radius R (w/o having to solve for "y") is:

    [tex]R=\frac{a^2+b^2}{2a}[/tex]

    where "a" is the change in y-intercept and "b" is the shift in the x direction.

    - Kamataat
     
  9. May 22, 2005 #8
    Kamataat, y is the radius in the formula (y - 5)^2 + 45^2 = y^2. Your formula is simplified, but equivalent.
     
  10. May 23, 2005 #9
    Aaaaaaaaargh, I figured something was not right. Anyway, won't make the same mistake of posting when tired again.

    - Kamataat
     
  11. May 23, 2005 #10
    Indeed, a general formula for a situation like this may very well be

    [tex](r+\Delta y)^2 + (\Delta x)^2 = r^2, \Delta x < r/2[/tex]

    I'm not sure I can prove it, or whether it has already been proven, or whether it's trivial.
     
    Last edited by a moderator: May 23, 2005
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