Problem 1273

28.41. Let $R_{1}$ and $R_{2}$ be any pair of non-intersecting circles. The remaining four circles form a chain, so by the previous problem, the circles $S^{\prime}$ and $S^{\prime \prime}$, which touch $R_{1}$ and $R_{2}$ at the points of their intersection with the line of centers, intersect at right angles (Fig. 28.14). If $R_{2}$ lies inside $R_{1}$, then the radii $r^{\prime}$ and $r^{\prime \prime}$ of the circles $S^{\prime}$ and $S^{\prime \prime}$ are equal to $\left(r_{1}+r_{2}+d\right) / 2$ and $\left(r_{1}+r_{2}-d\right) / 2$, and the distance between their centers $d^{\prime}=2 r_{1}-r_{1}-r_{2}=r_{1}-r_{2}$. The angle between $S^{\prime}$ and $S^{\prime \prime}$ is equal to the angle between their radii drawn to the point of intersection, so $\left(d^{\prime}\right)^{2}=\left(r^{\prime}\right)^{2}+\left(r^{\prime \prime}\right)^{2}$ or, after transformations, $d^{2}=r_{1}^{2}+r_{2}^{2}-6 r_{1} r_{2}$.

In the case where $R_{1}$ and $R_{2}$ do not lie one inside the other, the radii of the circles $S^{\prime}$ and $S^{\prime \prime}$ are equal to $\left(d+\left(r_{1}-r_{2}\right)\right) / 2$ and $\left(d-\left(r_{1}-r_{2}\right)\right) / 2$, and the distance between the centers $d^{\prime}=r_{1}+r_{2}+d-\left(r_{1}^{\prime}+r_{2}^{\prime}\right)=r_{1}+r_{2}$. As a result, we get $d^{2}=r_{1}^{2}+r_{2}^{2}+6 r_{1} r_{2}$.