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Circle Equation

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data

    is:
    >= 1

    valid for a circle equation?
    See attached?

    2. Relevant equations

    reference attached

    3. The attempt at a solution

    reference attached
     

    Attached Files:

  2. jcsd
  3. Sep 24, 2009 #2

    rock.freak667

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    if you mean x2+y2≥1 , then yes it is a 'circle' in essence but the points (x,y) that satisfy the inequality do not lie on the circle itself like if if it was x2+y2=1
     
  4. Sep 26, 2009 #3

    tiny-tim

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    Hi DrMath! :smile:

    (have a ≥ and try using the X2 and X2 tags just above the Reply box :wink:)

    Your picture says |z - 2j| ≥ 1 …

    in other words, the magnitude of the "vector" from 2j to z has length ≥ 1 …

    so z is the exterior (and boundary) of the circle of length 1 and centre at 2j :smile:

    (btw, it's easier to use "vectors" rather than coordinates for as problem like this :wink:)
     
  5. Sep 26, 2009 #4
    Thanks tiny-tim
    oh.. how to become a Homework Helper here?
     
  6. Sep 27, 2009 #5

    tiny-tim

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    :wink: just help people with their homework! :smile:
     
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