Determine the ratio $k$ so that the sum of the series $\sum_{i=0}^{\infty} l_{i}$ is equal to the circumference of the outermost circle. Question:
Consider an infinite series of concentric circles, were the radii $r_{0}, r_{1}, r_{2}, ...$ form a geometric series with the ratio $k$, $0 < k < 1$.
From a point on the outermost circle, a tangent line is drawn to the circle just inside it, a tangent line from this circle to the one just inside it and so on.
The length of the successive tangents are $l_{0}, l_{1}, l_{2}, ...$

(the image is a bit grainy, but $r_{0}$ is the radius of the outermost circle)
Determine the ratio $k$ so that the sum of the series
$$\sum_{i=0}^{\infty} l_{i}$$
is equal to the circumference of the outermost circle.
Attemped answer:
According to the last part of the question:
$$\sum_{i=0}^{\infty} l_{i} = 2 \pi r_{0}$$
Since this is also a geometric series, it can be written as:
$$\sum_{i=0}^{\infty} l_{i} = \sum_{i=0}^{\infty} l_{0} k^{i} = l_{0} \frac{1}{1-k}$$
since $l_{0}$ is the first term and $k$ is the ratio. Putting these two together gives:
$$l_{0} \frac{1}{1-k} =  2 \pi r_{0}$$
However, this means that I have one equation and three unknowns. I suspect that there is something to be gotten from the fact that the circles are concentric and so this might yield a relationship between $l_{0}$ and $r_{0}$.
For instance, if we form two congruent triangles where $l_{0}$ and $r_{0}$ are two of the sides for the first one, and $l_{1}$ and $r_{1}$ two of the sides for the second one and use Pythagoras theorem, it looks like we get: 
$$l_{0} = \sqrt{r_{0}^{2} - r_{1}^{2}}$$
..but this just looks like we get another unknown into the mix instead.
Looks like I am a bit stuck. What are some productive ways to drive this question home?

Without loss of generality we may assume that the outer circle has radius $1$.
If $l_0$ is as in the OP, then $l_0^2=1-k^2$. So by scaling we have $l_1=k\sqrt{1-k^2}$, $l_2=k^2\sqrt{1-k^2}$, and so on.
The sum of the geometric series is $\frac{\sqrt{1-k^2}}{1-k}$, that is, $\frac{\sqrt{1+k}}{\sqrt{1-k}}$. Set this equal to $2\pi$ and solve for $k$.
