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    George's example, revisited
    From: Alexandre Eremenko
    Date: 2006 Jun 17, 05:36 -0400

    George's example of a "shifted circle" was the following.
    Begin with the circle centered at
    (Lat, Long)=(0,0)
    of radius 60 degrees.
    Shift every point 1 degree N.
    The resulting figure is not a circle,
    and the best fitting circle deviates by more than 14 miles from it.
    
    All this is correct.
    Indeed, choose 4 points on the original circle as follows:
    A=(0,60), B=(0,-60), C=(60,0) and D=(-60,0).
    They are shifted to the new points
    a=(1,60), b=(1,-60), c=(61,0) and d=(-59,0).
    Now consider the circle best fitting these four points.
    It is clear from symmetry that the center of this best
    fitting circle should have longitude 0.
    The equidistant point from c and d is x=(1,0).
    However, the distance from this point x to a (or to b)
    is about 50d46' (of a great circle). Simple trig is used to
    compute this.
    So the deviation of our figure from any circle is about 14
    miles, as George stated.
    
    This shows:
    a) I was wrong when I wrote that the "size of the circle
    is irrelevant, but only relevant is its distance from
    the poles". Size of the circle is also relevant.
    b) It is a bad idea to use position circles of large radius
    for running fixes. In any case, it is wrong to think that
    a shift of a position circle will be close to a circle.
    c) The only case when the use of position circle
    (as opposite to a position LINE on a Mercator chart) is
    justified is when the circle is small.
    
    Thus all statements of George on the subject (except that one cannot draw
    a circle through 3 points) were correct.
    
    Alex.
    
    
    

       
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