Construction: External and Internal Tangents

Suppose that we have two circles, $\Omega_1$ and $\Omega_2.$ We are asked to construct their common external and common internal tangents. Here is a step-by-step process explaining how to do this:

1. Construct the centers of $\Omega_1$ and $\Omega_2.$ Call them $O_1$ and $O_2$ respectively. Draw the line $O_1O_2.$

2. Construct the diameters of $\Omega_1$ and $\Omega_2$ perpendicular to $O_1O_2.$ Call these diameters $AB$ and $CD,$ respectively, such that $A$ and $C$ are on the same side of $O_1O_2.$

3. Draw lines $AC$ and $BD.$ These two lines, along with $O_1O_2,$ will intersect at a single point. Call this point $P.$

4. Draw lines $AD$ and $BC.$ These two lines, along with $O_1O_2,$ will intersect at a single point. Call this point $Q.$

5. Draw the tangent lines to $\Omega_1$ and $\Omega_2$ through $P.$ These are the common external tangents.

6. Draw the tangent lines to $\Omega_1$ and $\Omega_2$ through $Q.$ These are the common internal tangents.

Note that this process is only for the case in which $\Omega_1$ and $\Omega_2$ do not intersect and are not the same size. There are a few special cases, but they are all easy to handle:

If $\Omega_1$ and $\Omega_2$ intersect at two points and are not the same size, then the procedure for constructing the common external tangents is the same, and there are no common internal tangents.

If $\Omega_1$ and $\Omega_2$ are tangent at a point $X$ and are not the same size, then the procedure for constructing the common external tangents is the same, and the common internal tangent is simply the line that passes through $X$ and is perpendicular to $O_1O_2,$ which is easy to construct.

If $\Omega_1\cong\Omega_2,$ then the procedure for constructing the common internal tangent doesn't change from what was described above (it just depends on the number of points of intersection of the circles). However, to construct the common external tangents, one simply has to draw the lines $AC$ and $BD$ as described above (so in fact, there are even fewer steps!).

So why does this work? Well, when we construct the points $P$ and $Q$ as described above, we are really just constructing the centers of homothety of the two circles. The center of the homothety with a positive scale factor is the point of intersection of the common external tangents, and the center of the homothety with a negative scale factor is the point of intersection of the common internal tangents. The special cases for which this does not apply are easy enough to handle on their own.