I was fiddling with a question that has already been answered on MSE. I (ostensibly) came up with a different solution that was not listed for my own recreational fun. I am having trouble rigorously justifying (to myself) that one step in my solution is valid.
Original question [also provided in the linked question]: Consider a circle of radius one centered at (0,0) and a circle of radius 1/2 centered at (3/2,0). There are three lines that lay tangent to both circles, one is trivial and the other two are reflections of each other. Find the equation of one of the non-trivial lines.
My solution: I seek the equation of the line with a negative slope (laying atop the two circles).
part 1: When I look at a graph of the two circles and the tangent line, I am curious what is this lines $x-$ intercept? I imagine; what if there were more shrinking circles instead of just the first two - another with radius 1/8, 1/16, 1/32... each touching the one before it in a single point. This would look like this (though with the circles continuing indefinitely):
Each of these circles should also lay tangent to the solution line, and the total sum of the length of the circles in the positive $x$ direction should be converging to the intercept.
This length is easily found to be $$1+2 \cdot \sum_{k = 1}^{\infty} \left(\frac{1}{2}\right)^k = 3. $$
part 2: An equation for a line that lies tangent to the first circle at an arbitrary point in the first quadrant can easily be found to be $$f_a(x) = \frac{1-ax}{\sqrt{1-a^2}}.$$
Using the intercept from part 1 we can solve to yield $a = 1/3$ which allows us to determine the equation of the line.
How can I rigorously justify that the sum of the length of the circles will converge to the intercept of the line?
