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Math Help - Equation of a circle : Intersection of lines

  1. #1
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    Question Equation of a circle : Intersection of lines

    The problem is:

    Find the equation of a circle given its center (2, -3) and pass through the intersection of lines x - 2y = 6 and x + 3y = 1
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  2. #2
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    Quote Originally Posted by FailCalculus View Post
    The problem is:

    Find the equation of a circle given its center (2, -3) and pass through the intersection of lines x - 2y = 6 and x + 3y = 1
    1. find the point of intersection of the lines (solve the simultaneous equations).

    2. find the distance from the point of intersection found above to the point (2,-3), this is the radius.

    3. You now have the centre and radius, so you can write down the equation.

    If you have any problems let us know what they are and we will provide help specific to your difficulties.

    CB
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  3. #3
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    x - 2y = 6
    x = 2y + 6 ...(1)
    x + 3y = 1
    x = 1 - 3y ...(2)

    2y + 6 = 1 - 3y
    x = 4 and y = -1
    Hence (4, -1) is a point on the circle with centre (2, -3) and radius r.
    r^2 = (4 - 2)^2 + (-1 - 3)^2
    r^2 = 8

    (y + 3)^2 + (x - 2)^2 = 8
    x^2 + y^2 - 4x + 6y + 5 = 0
    Last edited by ukorov; November 15th 2009 at 12:40 AM.
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