Suppose you would like to find the two tangent lines that support two given disks in the plane to the same side. Parameterizing the circles using $( \cos \theta, \sin \theta )$, I find that ultimately I am computing the roots of a 4th-degree polynomial in $x = \cos \theta$ (two of whose roots are imaginary, and two real). But I may not have formalized this in the most perspicacious manner. Rather than detail my approach, let me just pose the general question of whether there is a way to compute the tangent lines without effectively finding the roots of a 4th-degree polynomial. With two solutions one might hope for a 2nd-degree polynomial.
Addendum. I should have made clear that what I need are the coordinates of the two points of tangency for two circles arbitrarily placed in the plane. Nevertheless, both Ross's and Moron's solutions can answer this version of the question as well. With appropriate translation, scaling, and rotation, I can place the two circles so that one is a unit-radius circle centered on the origin, and the other has $r \ge 1$ with its center on the $+x$ axis. Then I am in the situation Moron drew. Knowing the side lengths of the triangles yields $\sin \theta$ where $\theta$ is the angle of the triangle on the $x$-axis, and from that I can compute the coordinates of the points of tangencies in terms of $\sin \theta$ and $\cos \theta$. Much simpler than my brute-force calculation. Thanks!