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Math Help - equation of circle

  1. #1
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    equation of circle

    the question i have is: the points A(2, 5) and B(-4, 13) lie at opposite ends of a diameter of a circle with centre C

    a) find the radius of the circle- i did this by finding the diameter which was 10 and halving it, so the radius is 5

    b) (which is the one i had a problem with): find the equation of the circle in the form x^2 + y^2 + px + qy + c = 0

    i found that the centre was (-1, 9) like so \frac{2 + (-4)}{2} = -1 and \frac {5 + 13}{2} = 9

    so the equation of the circle would be this: (x + 1)^2 + (y - 9)^2 = 25

    the answer the book gave was x^2 + y^2 + 2x - 18y + 57 = 0 which would imply that you made the squared constant terms positive, but it thought you would make them negative ie (x + 1)^2 - 1 + (y - 9)^2 - 81 = 25

    does the method i used only work for certain things?

    thanks
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  2. #2
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    Quote Originally Posted by mark View Post
    the question i have is: the points A(2, 5) and B(-4, 13) lie at opposite ends of a diameter of a circle with centre C

    a) find the radius of the circle- i did this by finding the diameter which was 10 and halving it, so the radius is 5

    b) (which is the one i had a problem with): find the equation of the circle in the form x^2 + y^2 + px + qy + c = 0

    i found that the centre was (-1, 9) like so \frac{2 + (-4)}{2} = -1 and \frac {5 + 13}{2} = 9

    so the equation of the circle would be this: (x + 1)^2 + (y - 9)^2 = 25

    the answer the book gave was x^2 + y^2 + 2x - 18y + 57 = 0 which would imply that you made the squared constant terms positive, but it thought you would make them negative ie (x + 1)^2 - 1 + (y - 9)^2 - 81 = 25

    does the method i used only work for certain things?

    thanks

    Just open up parentheses in the circle's equation!

    (x+1)^2+(y-9)^2=81\Longrightarrow x^2+2x+y^2-18y+82=25...

    Tonio
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