Let the circles be positioned at points p1, p2, p3 with radii r1, r2, r3 respectively. Note that we can get a family of circles with the same exterior tangents

Let the circles be positioned at points p1, p2, p3 with radii r1, r2, r3 respectively. Note that we can get a family of circles with the same exterior tangents by simultaneously inter/extra-polating the position and radius of two other circles. Solving for a radius of 0 gives the following position for the intersection of those the tangents of circles 1 and 2. q3 = (r1 * p2 - r2 * p1) / (r1 - r2) (with similar expression for q1 and q2). Using homogenous coordinates we get the following list for q1, q2, q3: ( r2 p3 - r3 p2 ) ( r3 p1 - r1 p3 ) ( r1 p2 - r2 p1 ) ( r2 - r3 ), ( r3 - r1 ), ( r1 - r2 ) now note that r1 q1 + r2 q2 + r3 q3 = 0, showing that they're not linearly independent and hence collinear. A somewhat neater but more advanced proof, follows by making the positions pi also homogeneous and noting that (using Einstein summation notation): 0 = ε_ijk rj rk so we can determine the intersections to be qi = ε_ijk rj pk hence ri qi = ε_ijk ri rj pk = ε_jik rj ri pk = -ε_ijk ri rj pk = 0.