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General Form of equation for a Circle

  1. Jul 24, 2007 #1
    I was helping my girlfriend earlier with some HW and now I don't have her text (since she is not here) but I have a question regarding circles.
    I did not cover a lot of this material in precalc. and only recall the Standard Form of an equation: x^2+y^2=r^2.

    So my question is: if given the equation in General Form, and I wish to know the radius, is the following procedure (completing the square) correct?

    Given x^2+y^2+ax+by+c=0

    x^2+ax+y^2+by+c=o
    x^2+ax+(a/2)^2+y^2+by+(b/2)^2+c=(a/2)^2+(b/2)^2
    (x+a/2)^2+(y+b/2)^2=(a/2)^2+(b/2)^2-c

    so r=sqrt((a/2)^2+(b/2)^2-c)

    is this correct?
    thanks,
    Casey
     
  2. jcsd
  3. Jul 24, 2007 #2

    CompuChip

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    Homework Helper

    Yep, it is correct - and a nice way to do it too.

    Another way to derive this is as follows: the equation of a circle with radius r and center point ([itex]x_0[/itex], [itex]y_0[/itex]) is [tex](x - x_0)^2 + (y - y_0)^2 = r^2[/tex]. Working out the brackets gives [tex]x^2 - 2 x_0 x + x_0^2 + y^2 - 2 y_0 y + y_0^2 - r^2 = 0[/tex]. Now if we compare to your formula [tex]x^2 + y^2 + a x + b y + c = 0[/tex] we read off that
    [tex]a = -2 x_0, b = 2 y_0, c = x_0^2 + y_0^2 - r^2[/tex]
    whence solving for r gives
    [tex] r^2 = x_0^2 + y_0^2 - c [/tex] and you get the same formula (you can substitute -a/2 for [itex]x_0[/itex] and drop the minus since it's squared anyway, and similarly for -b/2).
     
  4. Jul 24, 2007 #3
    Nice. Thank you CompuChip.
     
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