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Equations of Circles

  1. Oct 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Here is the equation of a circle:
    x^2 + y^2 - 3x + 7y - 6 = 0


    It is your job to find the center and the radius of the circle.

    --------------------------------------------------------------------------------

    Enter the coordinates of the center and the radius (as numbers):


    2. Relevant equations

    (x-a)^2+(y-b)^2=r^2
    (a,b)=center


    3. The attempt at a solution
    The problem: x^2 + y^2 - 3x + 7y - 6 = 0
    (x-3/2)^2+(y-(-7/2)^2

    (x-3/2) (x-3/2)= x^2 -3/2x -3/2x +9/4= x^2-3x+9/4
    (y+7/2) (y+7/2)= y^2 +7/2y +7/2y +49/4= y^2+7y+49/4

    So, x^2+y^2-3x+7y-29/2= -sqrt(17/2)

    I am doing independent study on a computer program. I have the values for the x and y coordinates right, but my value for the radius (sqrt(17/2)) is not right. I do not understand this because when I square this radius and move it over to the other side I get -6 just like the original equation of x^2 + y^2 - 3x + 7y - 6 = 0. What am I doing wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 21, 2010 #2

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    Try arranging your work like this:

    x2 - 3x + y2 + 7y = 6

    Now add to both sides:

    x2 - 3x + 9/4 + y2 + 7y + 49/4= 6 + 9/4 + 49/4

    (x - 3/2)2 + (y + 7/2)2 = 82/4 = 41/2

    Now you can read off the center and radius.
     
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