Tangent to two disks: Roots of a 4th-degree polynomial? 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!

I hope I understood the question correctly.
Consider the following (sorry for the crude image):

Using similar triangles we get that
$$\frac{D'+D}{D'} = \frac{R}{r}$$
We can easily solve for $D'$.
$r,R$ are the radii of the circles and and $D$ is the distance between the centres of the circles.
$D'$ is the distance between the common point of the tangents to the center of the smaller circle.
