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!

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You mean tangent lines? Of course the two disks cannot be concentric. –  Ｊ. Ｍ. Oct 10 '10 at 2:30
J.M.: Yes, thanks, I edited it to "tangent lines." And yes, there are degenerate cases. –  Joseph O'Rourke Oct 10 '10 at 2:35
One pretty simple case is the case of two externally tangent disks; the radical line is the tangent line at the point of mutual tangency. –  Ｊ. Ｍ. Oct 10 '10 at 2:40
Another use of the radical line: if it intersects both disks, there are only two tangents; if not, four. (I exclude the case where you have one small disk entirely inside one big disk, and the disks are not concentric.) –  Ｊ. Ｍ. Oct 10 '10 at 2:46
Ah, I should have mentioned this first... –  Ｊ. Ｍ. Oct 10 '10 at 3:01

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.

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"We can easily solve for D'" – yes, D' is Dr/(R-r). –  ShreevatsaR Oct 10 '10 at 5:31
Note how the difference of the radii appears here, showing this is essentially the same as Ross's observation. Great! –  Joseph O'Rourke Oct 10 '10 at 15:42

The degree 4 polynomial appears because the circles have 4, not 2, tangent lines algebraically: two internal and two external. In the question, the phrase "tangent lines that support" singles out the external tangents, which are lines of support, but this is not a distinction that can be made algebraically given a coordinatization of the problem. The subtlety is that the equations of the circles contain only $r^2$ and $R^2$ while a choice of which pair of tangents lines counts as "internal" and which pair is "external" requires a choice of sign for $r/R$ to determine the relative orientation of the circles.

The internal and external pairs of tangent lines to two circles can be constructed by ruler and compass, using one intersection of circles in each case, which in coordinates is solution of a quadratic equation. The diagram posted in the other answer, with $(R-r)$, is the same as the one for the Euclidean construction. The quartic with roots corresponding to the tangent lines is a product of two quadratics $Q_{int}(x)Q_{ext}(x)$ for the internal and external construction problems.

Among the 4 tangent lines the difference between the internal and external pairs can be seen algebraically, as choices of sign for different square roots. This naturally partitions the set of 4 roots into two pairs. However, a further labelling of the pairs as "internal" and "external" requires a further choice beyond a coordinatization of the plane. This is because the coefficient field of the Cartesian equations defining the tangent lines is $k(d_x,d_y,R^2,r^2)$ where $d_x$ and $d_y$ are the $x,y$ distance between centers and $k$ is the underlying field of coordinates, but the selection of which pair of tangents is external relies on a value of $r/R$ or $(R-r)^2$ and this requires some sign choices for the radii, or equivalently, an extension of the coefficient field by some square roots.

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In the wikipedia article tangent lines to circles there is a construction that appears to be second order. You construct the tangents from the center of the smaller circle to a circle, centered on the center of the larger, with radius the difference of the radii. Your tangents are parallel to these, offset by the smaller radius.

Sorry for the broken link-an extra / got in the way

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That link takes me to "Wikipedia does not have an article with this exact name." But just your description suffices. Thanks! –  Joseph O'Rourke Oct 10 '10 at 14:35