There are two possible configurations as shown in the diagrams below.


Let $O$ be the intersection point of $L_1$ and $L_2$. The centre $P$ of circle $C$ lies on the bisector of the angle between $L_1$ and $L_2$. Because $C_1$ and $C_2$ have equal radius, the line $OP$ is the common tangent of these two circles. The angle between $OP$ and $L_1$ is bisected by $OR$, where $R$ is the centre of $C_1$.

We now work with coordinates in which $L_1$ is the x-axis, $L_2$ the y-axis and $P$ has coordinates $(r, r)$. Let $R$ have coordinates $(x, y)$. Then $y = s$ is the radius of the circles $C_1$ and $C_2$ and $|PR|^2 = (x - r)^2 + (y - r)^2$. Since $C$ and $C_1$ are externally tangent to each other, $|PR| = r + s = r + y$ and so $(r + y)^2 = (x - r)^2 + (y - r)^2$ which simplifies to $(x - r)^2 = 4ry$.
As seen above, the angle between $L_1$ and the line $OR$ is a quarter of a right angle, hence the slope of $OR$ is equal to $m = \tan(\pi/8)$, and $y = mx$. Thus, $(x - r)^2 = 4rmx$. This expands as
$$x^2 - 2r(1 + 2m)x + r^2 = 0$$
with solutions
$$x_{1,2} = r(1 + 2m) \pm r\sqrt{(1 + 2m)^2 - 1} = r(1 + 2m \pm 2\sqrt{m^2 + m})$$
The two possibilities for $s$ are then
$$s_{1,2} = mx_{1,2} = rm(1 + 2m \pm 2\sqrt{m^2 + m})$$

Alternatively, we can use basic trigonometry to find a value for $m$. The half-angle formula for the tangent function says $\tan(\alpha/2) = (1 - \cos(\alpha))/\sin(\alpha)$. In our case, $\alpha = \pi/4$ and $\cos(\pi/4) = \sin(\pi/4) = 1/\sqrt{2}$. Hence
$$m = \tan(\pi/8) = \frac{1 - \cos(\pi/4)}{\sin(\pi/4)} = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \sqrt{2} - 1.$$

The smaller value corresponds to the diagram on the right, the larger to the diagram on the left at the start of this solution.
We can find a value for the slope $m$ of $OR$ with the aid of the Angle Bisector Theorem which says that an angle bisector divides the opposite side in a triangle in a ratio that is equal to the ratio between the sides adjacent to the angle. To apply this, we consider a right angled isosceles triangle $OLM$ with $|OL| = |LM| = 1$. If the bisector of $\angle LOM$ meets $LM$ at $R$, then $m = |LR|$ is the slope of $OR$.

We now obtain $m^2 + m = m(m + 1) = (\sqrt{2} - 1)\sqrt{2} = 2 - \sqrt{2}$ and so
$$s_{1,2} = r(\sqrt{2} - 1)\left(2\sqrt{2} - 1 \pm 2\sqrt{2 - \sqrt{2}}\right).$$