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Center of circle from two points and a tangent angle

  1. Feb 15, 2012 #1

    bkx

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    So my problem is this: I need to figure out the center of a circle given two points. At one of the points, I know the tangent angle. So I know (x1, y1, θ1) and (x2, y2) and need to find (xc, yc). I also need to do this on a computer so I need some sort of closed-form solution.

    The way I have approached this so far is to construct a line with the equation

    x(t) = x1 + t*cos(θ+pi/2)
    y(t) = y1 + t*sin (θ+pi/2)

    This line runs through (x1,y1) as well as (xc, yc). There is some value of t (equal to + or - the radius of the circle) where (x(t),y(t)) = (xc,yc). Given that center of the circle is equidistant to both (x1,y1) and (x2,y2), I equate the squared distances:

    [x(t)-x1]^2 + [y(t)-y1]^2 = [x(t)-x2]^2 + [y(t)-y2]^2

    I substitute in equations for x(t) and y(t) and use a CAS to solve for t, although I come up with an extremely long, convoluted expression.

    I have the feeling that there's got to be a simple, elegant way to do this, although I'm just not seeing it. Can anyone provide some insight or suggestions on how to approach this?

    Thanks!
     
  2. jcsd
  3. Feb 15, 2012 #2

    bkx

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    Solved it! Found a theorem stating that the tangent-chord angle is twice the angle of the arc. With the arc angle, I can solve for the radius by r = chord_length/(2*sin(arcangle/2)) and the rest is trigonometry.
     
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