1. On the plane, there are two non-intersecting circles of equal radius and a point $O$ - the midpoint of the segment with endpoints at the centers of these circles. A line $l$, parallel to the line of centers of these circles, intersects them at points $A, B$, $C$ and $D$. A line $m$, passing through $O$, intersects them at points $E, F, G$ and $H$. Find the radius of these circles, given that $A B=B C=C D=14$ and $E F=F G=G H=6$.

![](https://cdn.mathpix.com/cropped/2024_06_04_7a3ed3d68324df2727b5g-03.jpg?height=1011&width=1947&top_left_y=57&top_left_x=60)

Fig. 1: to the condition of problem 2
1. Let $T_{1}, T_{2}$ be the projections of the centers $O_{1}$ and $O_{2}$ of the circles onto the line $A D$. Then $T_{1}$ is the midpoint of $A B$, and $T_{2}$ is the midpoint of $C D$, so $T_{1} T_{2}=28$. Since $O_{1} T_{1} T_{2} O_{2}$ is a rectangle, $O_{1} O_{2}=28$, and by symmetry $O O_{1}=14$. Drop a perpendicular $O_{1} M$ from $O_{1}$ to the line $E H$. Then $M$ is the midpoint of $E F$, and $M F=3$. By symmetry, $O F=F G / 2=3$, $M O=M F+O F=6$. By the Pythagorean theorem for $\triangle M O O_{1}$, we get $O_{1} M^{2}=O O_{1}^{2}-M O^{2}=14^{2}-6^{2}=160$. By the Pythagorean theorem for $\triangle M F O_{1}$, we get $O_{1} F^{2}=O_{1} M^{2}+M F^{2}=160+9=169$. Therefore, the radius of the circles is 13.