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.

I studied it at the University, it was introduced after Pappo's theorem and other theorem demonstrated using Descriptive Feometry most of all.

On the other end it was a Projective Geometry course and at the end we used a lot the General Linear Group, in particular 4x4 matrices for bilinear forms, so it was like warming up with Geometry to arrive to the Algebra tools.

He's Pappus in English, for what it's worth.