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Circle eqauation - can ne1 double check my work

  1. Apr 20, 2004 #1
    Find the center and radius of the circle
    16x^2 + 16y^2 + 8x + 32y + 1 = 0

    So first i simplified the equation by taking out the 16
    so i got:

    16 (x^2 + 1/2x + y^2 + 2y) = -1
    16 (x^2 + 1/2x + 1/16) + 16 (y^2 + 2y + 1) = -1 + 1 + 1
    16 (x + 1/4)^2 + 16 (y + 1)^2 = 1

    Center = (-1/4, -1)
    Radius = 1

    Are my calculations correct? Do I need to take out the 16 in my equation?
     
  2. jcsd
  3. Apr 21, 2004 #2
    I believe you should write it in the form (x + 1/4)^2 + (y + 1)^2 = 1/16 = (1/4)^2. Then you can see that the radius is 1/4.
     
  4. Apr 21, 2004 #3
    uh oh, then I must have done something wrong because the answer for the Radius should be just 1 :frown:
     
  5. Apr 21, 2004 #4

    Integral

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    On the left hand side you added 16 on the right hand side you added 1.
    If you add 16 to both sides it will work out.
     
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