Equation of the line that lies tangent to both circles Consider the two circles determined by $(x-1)^2 + y^2 = 1$ and $(x-2.5)^2 + y^2 = (1/2)^2$. Find the (explicit) equation of the line that lies tangent to both circles.

I have never seen a clean or clever solution to this problem. This problem came up once at a staff meeting for a tutoring center I worked at during undergrad. I recall my roommate and I - after a good amount of time symbol pushing - were able to visibly see a solution by inspection, then verify it by plugging in. I have never seen a solid derivation of a solution to this though, so I would like to see what MSE can come up with for this!

I took a short stab at it today before posting, and got that it would be determined by the solution to the equation $$\left( \frac{\cos(\theta)}{\sin(\theta)} + 2\cos(\theta) + 3 \right)^2 -4\left( \frac{\cos(\theta)^2}{\sin(\theta)^2}+1 \right)\left(\frac{-\cos(\theta)^3}{\sin(\theta)^2}+\frac{\cos(\theta)^4}{\sin(\theta)^2}+3 \right).$$
The solution $\theta$ would then determine the line $$y(x) = \frac{-\cos(\theta)}{\sin(\theta)}(x) + \frac{\cos(\theta)^2}{\sin(\theta)} + \sin(\theta).$$
Not only do I not want to try and solve that, I don't even want to try expanding it out :/

As peterwhy points out in the comments, there are three tangent lines.  By inspection, one is $x = 2$, as pointed out by J.W. Tanner in the comments.
The other two can be identified by similar triangles.  Suppose that we have a line tangent to both circles, and let the points of tangency be $T_1$ and $T_2$.  Let the circle centers be $O_1$ and $O_2$.  Finally, let the point where this line intersects the $x$-axis be called $P$.  Then $\triangle PO_1T_1$ and $\triangle PO_2T_2$ are similar (do you see why?).  Since $O_1T_1 = 2O_2T_2$, we must have $PO_1 = 2PO_2$, and therefore $P$ must be at $(4, 0)$.  Note that $PT_1 = \sqrt{3^2-1^2} = \sqrt{8}$, and therefore our tangent line must have slope $\pm \frac{1}{\sqrt{8}}$.

(For simplicity, I only show one of the tangent lines; the other is its mirror image across the $x$-axis.)  From this, we get the equation of the two remaining tangent lines
$$
y = \pm \frac{x-4}{\sqrt{8}}
$$
