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: Centre Point of a circle

  1. Oct 31, 2007 #1
    URGENT: Centre Point of a circle

    1. The problem statement, all variables and given/known data
    Find the center point of the circle shown in the diagram below and the radius of the circle.


    The red dots have the following coordinates; (-17,0); (0,17); (31,0)

    2. Relevant equations
    [tex]\left( {x + a} \right)^2 + \left( {y + b} \right)^2 = r^2 [/tex]

    3. The attempt at a solution
    I can find the x-coordinate of the center of the circle finding the center point between (-17,0) and (31,0), and as the circle is symmetrical, this will be the x-coordinate of the center point. The value is 7, therefore the center of the circle is located (7,c) where c is some constant.

    how do i find the value of the constant?

    Many thanks,
  2. jcsd
  3. Oct 31, 2007 #2


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    Homework Helper

    The distance between (7,c) and (0,17) equals the distance between (7,c) and (31,0).

    use this to find c, and and then r also...
  4. Oct 31, 2007 #3
    sorry i don't see how that helps

    would simultaneous solving of some sort help?
    Last edited: Oct 31, 2007
  5. Oct 31, 2007 #4


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    Staff Emeritus
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    The perpendicular bisector of any chord in a circle goes through the center. You used that when you calculated the midpoint of the horizontal line between (-17,0) and (31,0): its midpoint is (7,0) and its perpendicular bisector is x= 7. Now choose another two of those points, say (-17, 0) and (0, 17). The midpoint of that segment is (-17/2, 17/2). The slope of that segment is (17-0)/(0-(-17))= 1. The slope of a line perpendicular to that is -1. The equation of the perpendicular bisector of that segment is y= -x+ 17. The center is at the intersection of y= -(x+ 17/2)+ 17/2 and x= 7. That should be easy to solve.
  6. Oct 31, 2007 #5
    So the centre is (7, c).
    Length of segment joining (-17, 0) and center
    r^2 = 24^2 + c^2
    r^2 = 576 + c^2 _____(Eq 1)
    Length of segment joining (0, 17) and center
    r^2 = 7^2 + (c-17)^2
    r^2 = 49 + c^2 + 289 -34c _______(Eq 2)

    Comparing Eqs. 1 and 2:
    576 + c^2 = 49 + c^2 + 289 -34c
    c = -7

    Therefore the centre of the circle is (7, -7).

    Finding radius now is pretty trivial.
  7. Oct 31, 2007 #6
    cheers for that, makes sense now, the radius is [tex]\sqrt {149} [/tex]? correct...
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