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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!
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!