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Analytic Geometry Question (equation of a circle)

  1. Jul 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the equation of the circle whose centre lies on the x axis and which passes through points A (6,0) and B (0,10).

    2. Relevant equations



    3. The attempt at a solution

    I drew a diagram of the circle and determined that the line AB has gradient 5/3. Its perpendicular bisector should pass through the centre of the circle. This perpendicular line has a gradient of -3/5 and, I think, an equation of 3x+5y-34=0.

    The centre lies on the x-axis, so the y-coordinate at the centre must be 0. Hence I can use the equation of the perpendicular line to solve for x at point (x,0). I get 34/4 for x here, giving a centre at (34/3 ,0).

    I am not sure where to go now, or if the above is even correct. Any help greatly appreciated.
     
    Last edited: Jul 7, 2012
  2. jcsd
  3. Jul 7, 2012 #2

    eumyang

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    The slope is -5/3, not 5/3. :wink: So the slope of the perpendicular will instead be 3/5, and the equation of the line will be... ?
     
  4. Jul 7, 2012 #3
    Whoops...I'll blame that one on lack of sleep! ;)

    So I'm on the right track? The equation of the perpendicular line is now (I hope) 3x-5y-16=0.
     
  5. Jul 7, 2012 #4

    eumyang

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    I got 3x - 5y + 16 = 0. :wink:

    After you do this, find the distance from the center to either A or B. That will give you the radius. Once you have the center and radius, you can write the equation of the circle.
     
  6. Jul 8, 2012 #5

    ehild

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    You can find the centre of the circle without much geometry, just using the fact that the given points are at equal distance R from it. If the coordinates of the centre are (a,0), a2+102=(6-a)2=R2.

    ehild
     
  7. Jul 9, 2012 #6
    Thanks for your replies ehild and eumyang, you were a great help.
     
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