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If it is 0

  1. Feb 22, 2005 #1
    if it is 0 ....

    hey all ....
    if we have the circle equation and we found the radius .... if it is positive the circle would off course be real ....... but if it was 0, do we say that we don't have a circle or the circle is a dot ?????
    cheers
    abc
     
  2. jcsd
  3. Feb 22, 2005 #2

    HallsofIvy

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    The graph is a single point. Technically, that's NOT a circle but you can consider it a "degenerate" circle.

    (and if r< 0, your graph is the empty set- some people consider that a REALLY degenerate circle!)
     
  4. Feb 22, 2005 #3

    dextercioby

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    "Technically",the equation would be (for one with the center in the origin)
    [tex] x^{2}+y^{2}=R^{2} [/tex] and the radius being negative would not mean anything wrong.The pairs (x,y) satisfying it would be the same.As for "zero",well,that should be seen as the "degenerate" case...

    The wrong part with negative radius would come from the definition of the circle,which uses the word "distance",and the definition of distance for a metric space (in this case [itex] R^{2} [/itex]) which is an application to [itex] R_{+}\cup \{0\} [/itex]..

    Daniel.
     
  5. Feb 22, 2005 #4
    One could argue that the circle in question exists in complex space where either or both of x and y are complex.

    Then again, if R^2 is negative the radius is also complex. does this invalidate the nature of a circle or is it reasonable?
     
  6. Feb 22, 2005 #5

    dextercioby

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    No,it does not invalidate the "nature" of the circle,it just pushes our power of understanding...The abstractization is sometimes unappropriate.The circle MUST have a real positive nonzero radius.And it's preferable to have as a one-dimensional object...

    Daniel.
     
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