5. The distance between the centers $O_{1}$ and $O_{2}$ of circles $\omega_{1}$ and $\omega_{2}$ is $5 r$, and their radii are $r$ and $7 r$ respectively. A chord of circle $\omega_{2}$ is tangent to circle $\omega_{1}$ and is divided by the point of tangency in the ratio $1: 6$. Find the length of this chord.

( 7 points)
Answer: $7 r \sqrt{3}$ or $\frac{7 r}{6} \sqrt{143}$.

Solution: Let $O_{1}$ be the center of the first circle $\omega_{1}$, and point $M$ be the point of tangency of this circle with the chord $A B$. From the center $O_{2}$ of the second circle $\omega_{2}$, draw $O_{2} K \perp A B$, where point $K$ is the midpoint of the chord $A B$. Drop $O_{1} L \perp O_{2} K$ and consider three different cases of the position of the chord $A B$ and the centers $O_{1}$ and $O_{2}$.

1) Suppose $O_{1}, O_{2}$ lie in the same half-plane relative to the line $A B$ and $O_{2} K \geq r$ (Fig.4). Since $K L=O_{1} M=r$, then $L$ lies on the segment $O_{2} K$. Let $A M=x$, then $A B=7 x, B K=7 x / 2, M K=O_{1} L=5 x / 2$. In the right triangle $\Delta O_{2} K B$, we have $O_{2} K=\frac{7}{2} \sqrt{4 r^{2}-x^{2}}$. In the right triangle $\Delta O_{1} O_{2} L$, we have $O_{1}, O_{2}=5 r, O_{1} L=\frac{5 x}{2}$, from which we get $O_{2} L=\frac{5}{2} \sqrt{4 r^{2}-x^{2}}$. Considering that $O_{2} L+K L=O_{2} K$, we get the equation $r=\sqrt{4 r^{2}-x^{2}}$. Solving it, we find $x=r \sqrt{3}$. Therefore, $A B=7 r \sqrt{3}$.
2) Suppose $O_{1}, O_{2}$ still lie in the same half-plane relative to the line $A B$ (Fig.5), but $O_{2} K<r$ (in particular, it may be $O_{2}=K$, that is, $A B$ is a diameter). Therefore, $L$ lies on the extension of the segment $O_{2} K$, and $O_{2} K+O_{2} L=K L$. The expressions for $O_{2} K, O_{2} L$, $K L$ in terms of $x$ and $r$ remain the same, and the equation for finding $x$ will take the form $r=6 \sqrt{4 r^{2}-x^{2}}$. Solving it, we find $x=r \frac{\sqrt{143}}{6}$. Therefore, $A B=7 r \frac{\sqrt{143}}{6}$.
3) Suppose $O_{1}, O_{2}$ lie on opposite sides relative to the line $A B$ (Fig.6). Therefore, $L$ lies on the extension of the segment $O_{2} K$, and $O_{2} K+K L=O_{2} L$. The expressions for $O_{2} K, O_{2} L$, $K L$ in terms of $x$ and $r$ remain the same, and the equation for finding $x$ will take the form $r+\sqrt{4 r^{2}-x^{2}}=0$ and has no solutions.

## MINISTRY OF SCIENCE AND HIGHER EDUCATION OF THE RF COUNCIL OF RECTORS OF UNIVERSITIES OF TOMSK REGION OPEN REGIONAL INTERUNIVERSITY OLYMPIAD 2018-2019 MATHEMATICS (10th GRADE) FINAL STAGE