The circle $k_{1}$ with center $O_{1}$ and radius 2 units, and the circle $k_{2}$ with center $O_{2}$ and radius $r$, touch each other externally. The Thales circle of the segment $O_{1} O_{2}$ intersects the circle $k_{1}$ at points $A$ and $B$, and the circle $k_{2}$ at points $C$ and $D$. The reflection of $O_{1}$ over the segment $A B$ is $O_{1}^{\prime}$, and the reflection of $O_{2}$ over the segment $C D$ is $O_{2}^{\prime}$. The distance $O_{1}^{\prime} O_{2}^{\prime}$ is 1 unit.

What is the value of $r$?
Apply the Pythagorean theorem to the right-angled triangles $O_{1} A O_{2}$ and $O_{1} C_{2}$ (1st, 2nd figure):

$$
O_{1} T_{1}=\frac{2^{2}}{O_{1} O_{2}}=\frac{4}{r+2} ; \quad O_{2} T_{2}=\frac{r^{2}}{O_{1} O_{2}}=\frac{r^{2}}{r+2}
$$

$1988-03-102-1 . \mathrm{eps}$

1st figure

$1988-03-103-2 . \mathrm{eps}$

2nd figure

Due to reflection,

$$
O_{1} O_{1}^{\prime}=2 O_{1} T_{1}=\frac{8}{r+2} \quad \text { and } \quad O_{2} O_{2}^{\prime}=2 O_{2} T_{2}=\frac{2 r^{2}}{r+2}
$$

When writing the length of the segment $O_{1}^{\prime} O_{2}^{\prime}$, we must take into account that the points $O_{1}, O_{1}^{\prime}, O_{2}^{\prime}, O_{2}$ do not necessarily lie in this order on the central line of the two circles; it is also possible that the reflected points fall outside the segment $O_{1} O_{2}$ (2nd figure). However, the vector equality

$$
\overrightarrow{O_{1}^{\prime} O_{2}^{\prime}}=\overrightarrow{O_{1}^{\prime} O_{1}}+\overrightarrow{O_{1} O_{2}}+\overrightarrow{O_{2} O_{2}^{\prime}}=\overrightarrow{O_{1}^{\prime} O_{1}}+\overrightarrow{O_{2} O_{2}^{\prime}}-\overrightarrow{O_{2} O_{1}}
$$

always holds. From the fact that $O_{1}^{\prime}$ is on the ray $O_{1} O_{2}$ and $O_{2}^{\prime}$ is on the ray $O_{2} O_{1}$, it follows that the vectors on the right side of (1), $\overrightarrow{O_{1}^{\prime} O_{1}}, \overrightarrow{O_{2} O_{2}^{\prime}}$ and $\overrightarrow{O_{2} O_{1}}$, are collinear (3rd figure).

# 1988-03-103-1.eps 

3rd figure

Then the length of the segment $O_{1}^{\prime} O_{2}^{\prime}$ can be expressed as the absolute value of the vector $\overrightarrow{O_{1}^{\prime} O_{2}^{\prime}}$ in the following way using the lengths of the vectors involved in the decomposition:

$$
\left|\overrightarrow{O_{1}^{\prime} O_{2}^{\prime}}\right|=O_{1}^{\prime} O_{2}^{\prime}=\left|O_{1}^{\prime} O_{1}+O_{2} O_{2}^{\prime}-O_{2} O_{1}\right|
$$

where on the right side within the absolute value, the lengths of the three segments stand. Using the given $O_{1}^{\prime} O_{2}^{\prime}=1$ (2), the following form can be written:

$$
1=\left|\frac{8+2 r^{2}}{r+2}-(r+2)\right|=\left|\frac{(r-2)^{2}}{r+2}\right|, \text { i.e., } 1=\frac{(r-2)^{2}}{r+2}
$$

From this, we get the equation $r^{2}-5 r+2=0$, whose roots are $\frac{5+\sqrt{17}}{2}$ and $\frac{5-\sqrt{17}}{2}$. Since our steps are reversible, both solutions are valid. The order of the points in the first case is $O_{1}^{2}, O_{2}^{\prime}, O_{1}^{\prime}, O_{2}$, and in the second case, $O_{1}, O_{2}^{\prime}, O_{2}, O_{1}^{\prime}$ (2nd figure).