circlef_t m1, m2;
linef_t   centers;
point2d_t  ppoint;

calcslope( &m1.c, &m2.c, &centers);
ppoint.y = (m1.c.y-m2.c.y)*(m1.r/m2.r)+m2.c.y;
ppoint.y = (m1.c.x-m2.c.x)*(m1.r/m2.r)+m2.c.x;
calefromslopepoint(&centers, &ppoint, &centers);
perpendicular(&centers);


y = mx + b
<wli> lhf: cos(A) = (d(a,b)^2 + d(a,c)^2 - d(b,c)^2)/(2*d(a,b)*d(a,c))
<wli> lhf: cos(B) = (d(b,a)^2 + d(b,c)^2 - d(a,c)^2)/(2*d(b,a)*d(b,c))
<wli> lhf: cos(C) = (d(c,a)^2 + d(c,b)^2 - d(a,b)^2)/(2*d(c,a)*d(c,b)) 


Let r = (r_x, r_y) 
d(w, r) = sqrt((w_x - r_x)^2 + (w_y - r_y)^2)

 d(b, c)^2 = d(r, b)^2 + d(r, c)^2 - 2*d(r, b)*d(r, c)*cos(LBRC)

http://img530.imageshack.us/img530/2268/step1ig8.jpg
http://img247.imageshack.us/img247/1320/step2em1.jpg







* klafka has quit ("Leaving")
<wli> rue_recovering: The normal methods of solving the system of algebraic equations for x and y would need you to do something like:
* tokj has quit ()
* rue_recovering tries to understand the answer
<wli> rue_recovering: 4*d(r, a)^2*d(r, b)^2 = (d(r, a)^2 + d(r, b)^2 - d(a, b)^2)^2, which is nasty. I would treat d(r, a), d(r, b), and d(r, c) as variables by themselves.