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Radius of circle, given 3 points

  1. Nov 2, 2009 #1
    1. The problem statement, all variables and given/known data
    The points (-6,2), (3,-1) and (-5,-5) are all points on circumference of the circle. What is the radius of this circle?


    2. Relevant equations
    I don't know what to do.



    3. The attempt at a solution
    Have no idea really. I tried actually making a makeshift graphical representation of it and still couldn't figure out how to to get the radius mathematically. The answer is r=5
     
  2. jcsd
  3. Nov 2, 2009 #2
    Use the equation for the circle:

    [tex](x-x_0)^2+(y-y_0)^2=r^2[/tex]

    Inserting the three points give you three equations with three unknows: The x and y coordinates of the cetnerpoint and the radius.
     
  4. Nov 2, 2009 #3

    LCKurtz

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    Call your points A, B, and C. An alternative approach is to find where the perpendicular bisectors of the chords AB and AC intersect. That would be the center. This only involves midpoints, slopes and equations of straight lines to work out.
     
  5. Nov 3, 2009 #4
    I was able to do it using LCKurtz method, but I wonder if using the circle formula would be faster. Can you help me out a little bit more? Can you give me one of the 3 equations I would make? I just need to know what to do with [tex]x_0[/tex] and the [tex]y_0[/tex]
     
  6. Nov 3, 2009 #5

    LCKurtz

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    It won't be easier. Call the unknown center (a,b). Your circle becomes:

    (x-a)2 + (y-b)2=r2.
    Plug in the first point, x = -6, y = 2:

    (-6-a)2 + (2-b)2 = r2
    36 +12a + a2 + 4 -4b + b2 = r2

    Plug in the other two points to get two more nonlinear equations like that to solve simultaneously.
     
  7. Nov 3, 2009 #6
    Though technically correct, nonlinear is a bit rash :) Quadratric equations are easy to solve.
     
  8. Nov 3, 2009 #7

    LCKurtz

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    Rash: (adj.) Hastily without due consideration. Done with reckless or ill-considered haste.

    Sorry, no. I meant what I said. Linear methods don't apply to a system of simultaneous quadratic equations like you get. My point was that that method is not easier than the direct method I suggested, and it isn't.
     
  9. Nov 4, 2009 #8
    Meant Harsh of course.

    Well I'd say id depends very much indeed on how you define "easy"

    Indeed, you can't solve a system of quadratic equations using a eg. gauss elimination, but for the linear approach, you still need to construct at least 4 additional "point/slope"-equations, so when it comes down to the actual amount of pencil carbon spent, I recon the two method are fairly evenly balanced.
     
  10. Nov 4, 2009 #9

    LCKurtz

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    I will concede that in this particular situation the system of equations you get using the distance formula happens to solve easily. Luckily, subtracting them eliminates all the squared terms, so the work is in fact approximately equal. But generally speaking, solving three equations in three unknowns involving both linear and quadratic terms is far from a walk in the park, and one would a priori expect the slope-straight line method to be easier and more direct.
     
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