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 $D E=\frac{48}{7}, D H=H C=\frac{1}{7}$. Find the length of segment $H P$ and the radii of both circles.
Answer: $H P=6, r=5 \sqrt{\frac{2}{3}}, R=5 \sqrt{\frac{2}{3}}$.

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

$$
\begin{gathered}
H F^{2}=H D \cdot H E=\frac{1}{7} \cdot 7=1 \Rightarrow H F=1 \\
H F^{2}=H C \cdot H B \Rightarrow H B=\frac{H F^{2}}{H C}=7 \\
B A^{2}=B D \cdot B E=\frac{50}{7} \cdot 14=100 \Rightarrow B A=10
\end{gathered}
$$

Since the segments of tangents drawn from a point to a circle are equal, $P F=P A=P B$, hence $P F=\frac{1}{2} A B=5$. Therefore, $P H=P F+F H=6$. 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., $49=25+36-60 \cos \gamma, \cos \gamma=\frac{1}{5}$.

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, $O Q=R+r, O A=R, B Q=r, A B=10, \angle A O Q=\gamma$. Dropping a perpendicular $Q H$ from point $Q$ to the base $A O$, we get a right triangle $O H Q$, in which $Q H=A B=10, O H=R-r$. By the Pythagorean theorem, we have $(R+r)^{2}=(R-r)^{2}+10^{2}$; also, $\frac{1}{5}=\cos \gamma=\frac{R-r}{R+r}$. From the last equation, we get $R=\frac{3 r}{2}$, and from the first, we get $R r=25$. Solving this system of equations, we find that $R=5 \sqrt{\frac{3}{2}}$, $r=5 \sqrt{\frac{2}{3}}$.