4. Circles $\Omega$ and $\omega$ touch each other externally at point $F$, and their common external tangent touches circles $\Omega$ and $\omega$ at points $A$ and $B$, respectively. Line $\ell$ passes through point $B$, intersects circle $\omega$ again at point $C$, and intersects $\Omega$ at points $D$ and $E$ (point $D$ is between $C$ and $E$). The common tangent of the circles passing through point $F$ intersects lines $A B$ and $B E$ at points $P$ and $H$, respectively (point $F$ lies between points $P$ and $H$). It is known that $D E=\frac{10}{3}, D H=H C=\frac{1}{9}$. Find the length of segment $H P$ and the radii of both circles.
Answer: $H P=\frac{5 \sqrt{31}}{9}, r=\frac{4 \sqrt{31}}{3 \sqrt{15}}, R=\frac{4 \sqrt{155}}{9 \sqrt{3}}$.

Solution. We apply the tangent-secant theorem three times:

\[
\begin{gathered}
H F^{2}=H D \cdot H E=\frac{1}{9} \cdot \frac{31}{9}=\frac{31}{81} \Rightarrow H F=\frac{\sqrt{31}}{9} \\
H F^{2}=H C \cdot H B \Rightarrow H B=\frac{H F^{2}}{H C}=\frac{31}{9} \\
B A^{2}=B D \cdot B E=\frac{32}{9} \cdot \frac{62}{9}=\frac{64 \cdot 31}{9^{2}} \Rightarrow B A=\frac{8}{9} \sqrt{31}
\end{gathered}
\]

Since the segments of tangents drawn from a point to a circle are equal, $P F=P A=P B$, hence $P F=\frac{1}{2} A B=\frac{4}{9} \sqrt{31}$. Therefore, $P H=P F+F H=\frac{5}{9} \sqrt{31}$. Let $\angle B P H=\gamma$. Then, by the cosine rule for triangle $B P H$, we have $B H^{2}=B P^{2}+H P^{2}-2 B P \cdot P H \cdot \cos \gamma$, i.e., $\left(\frac{31}{9}\right)^{2}=\left(\frac{4}{9}\right)^{2} \cdot 31+\left(\frac{5}{9}\right)^{2} \cdot 31-2 \cdot \frac{20}{9^{2}} \cdot 31 \cos \gamma$, from which $31=16+25-40 \cos \gamma$, $\cos \gamma=\frac{1}{4}$.

Let $O$ and $Q$ be the centers, and $R$ and $r$ be the radii of circles $\Omega$ and $\omega$ respectively; since the circles are tangent, the point of tangency $F$ lies on the line of centers $O Q$, and $O Q \perp P H$. Angles $A$ and $F$ of quadrilateral $A O F P$ are right angles, so $\angle A O F=180^{\circ}-\angle A P F=\angle B P F=\gamma$.

Consider the right trapezoid $A B Q O$. In it, $O Q=R+r, O A=R, B Q=r, A B=\frac{8}{9} \sqrt{31}, \angle A O Q=\gamma$. Dropping a perpendicular $Q H$ from point $Q$ to the base $A O$, we get a right triangle $O H Q$, in which $Q H=A B=\frac{8}{9} \sqrt{31}, O H=R-r$. By the Pythagorean theorem, we have $(R+r)^{2}=(R-r)^{2}+\left(\frac{8 \sqrt{31}}{9}\right)^{2} ;$ also, $\frac{1}{4}=\cos \gamma=\frac{R-r}{R+r}$. From the last equation, we get $R=\frac{5 r}{3}$, and from the first, we get $R r=\frac{16 \cdot 31}{81}$. Solving this system of equations, we find that $R=\frac{4 \sqrt{155}}{9 \sqrt{3}}, r=\frac{4 \sqrt{31}}{3 \sqrt{15}}$.