12. B6 (FIN 3) Four distinct circles \( C, C_{1}, C_{2}, C_{3} \) and a line \( L \) are given in the plane such that \( C \) and \( L \) are disjoint and each of the circles \( C_{1}, C_{2}, C_{3} \) touches the other two, as well as \( C \) and \( L \). Assuming the radius of \( C \) to be 1, determine the distance between its center and \( L \).
12. Let $y$ be the line perpendicular to $L$ passing through the center of $C$. It can be shown by a continuity argument that there exists a point $Y \in y$ such that an inversion $\Psi$ centered at $Y$ maps $C$ and $L$ onto two concentric circles $\widehat{C}$ and $\widehat{L}$. Let $\widehat{X}$ denote the image of an object $X$ under $\Psi$. Then the circles $\widehat{C_{i}}$ touch $\widehat{C}$ externally and $\widehat{L}$ internally, and all have the same radius. Let us now rotate the picture around the common center $Z$ of $\widehat{C}$ and $\widehat{L}$ so that $\widehat{C_{3}}$ passes through $Y$. Applying the inversion $\Psi$ again on the picture thus obtained, $\widehat{C}$ and $\widehat{L}$ go back to $C$ and $L$, but $\widehat{C_{3}}$ goes to a line $C_{3}^{\prime}$ parallel to $L$, while the images of $\widehat{C_{1}}$ and $\widehat{C_{2}}$ go to two equal circles $C_{1}^{\prime}$ and $C_{2}^{\prime}$ touching $L, C_{3}^{\prime}$, and $C$. This way we have achieved that $C_{3}$ becomes a line. Denote by $O_{1}, O_{2}, O$ respectively the centers of the circles $C_{1}^{\prime}, C_{2}^{\prime}, C$ and by $T$ the point of tangency of the circles $C_{1}^{\prime}$ and $C_{2}^{\prime}$. If $x$ is the common radius of the circles $C_{1}^{\prime}$ and $C_{2}^{\prime}$, then from $\triangle O_{1} T O$ we obtain ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-463.jpg?height=266&width=496&top_left_y=1028&top_left_x=832) that $(x-1)^{2}+x^{2}=(x+1)^{2}$, and thus $x=4$. Hence the distance of $O$ from $L$ equals $2 x-1=7$.