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Math Help - Minimum distance from a circle

  1. #1
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    Minimum distance from a circle

    Hi,
    I'm not sure if this is a very easy or a very hard question... What I know is that you have to be patient as my definition of it is probably not very rigorous.

    In a circle, the centre is equidistant from the points on the edge.
    However, can I also safely assume that the centre is the point at the minimum combined distance from the edge? In other words, is the sum of the distances from all the points on the edge of a circle to a certain point within the circle minimum if the given point is also the centre?

    Thanx for your answers.
    Matteo
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  2. #2
    Member Nacho's Avatar
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    I think if you want sum all lines from edge the circle to the point, you will obtain the superface of the circle. Hence donīt care the point you take, always the sum will be same
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  3. #3
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    Hi Matteo,

    I think your question makes sense if we interpret "sum of the distances from all the points on the edge of a circle to a certain point" as a suitable integral. For simplicity let's assume our circle has radius 1 and is centered at (0,0). Take arbitrary point P in R^2 and because our problem is invariant under rotation of coordinate axes we can assume P=(a,0), a >= 0. The intergral i have in mind is:

    \int_0^{2\pi}\sqrt{(a-\cos(x))^2+\sin^2(x)}\, \mbox{d}x = \int_0^{2\pi}\sqrt{1+a^2-2a\cos(x)}\, \mbox{d}x =\ldots

    this leads to a complete elliptic integral of the second kind, with reference to a \ge 0 we get

    \ldots = 4\cdot (a+1)\cdot E\left(2\sqrt{a/(a+1)^2}\right)

    and this has a minimum =2\pi at a=0.
    This means that point P=(0,0), which is the centre of the circle, is the point that minimizes the integral.
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  4. #4
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    Thanks

    Quote Originally Posted by Taluivren View Post
    Hi Matteo,

    I think your question makes sense if we interpret "sum of the distances from all the points on the edge of a circle to a certain point" as a suitable integral.
    ah, the wonderful tools of mathematics

    Quote Originally Posted by Taluivren View Post
    and this has a minimum =2\pi at a=0.
    This means that point P=(0,0), which is the centre of the circle, is the point that minimizes the integral.
    I'm a little late with this, but thank you very much!
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  5. #5
    Member Nacho's Avatar
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    Why my answer is wrong?

    Thanks
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