Center of circle from two points and a tangent angle

[bkx]

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!

 

[bkx]

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.