4. Points $A, B, C, D, E$ are sequentially located on a line, such that $A B=B C=2, C D=1, D E=3$. Circles $\Omega$ and $\omega$, touching each other, are such that $\Omega$ passes through points $A$ and $E$, and $\omega$ passes through points $B$ and $C$. Find the radii of circles $\Omega$ and $\omega$, given that their centers and point $D$ lie on the same line.
Answer: $R=8 \sqrt{\frac{3}{11}}, r=5 \sqrt{\frac{3}{11}}$.

Solution. Let the centers of circles $\Omega$ and $\omega$ be $O$ and $Q$ respectively. Since $C$ is the midpoint of chord $A E$ of circle $\Omega$, segment $O C$ is perpendicular to $O E$. Drop a perpendicular $Q H$ from point $Q$ to line $A E$. Then $B H=H C=1$ (a diameter perpendicular to a chord bisects the chord).

Let $O C=x$. Then $Q H=2 x$ (since $O C$ is the midline of triangle $D H Q$), $r=Q B=\sqrt{Q H^{2}+B H^{2}}=\sqrt{4 x^{2}+1}, R=O A=\sqrt{O C^{2}+A C^{2}}=\sqrt{x^{2}+9}$. We can express the segment $O Q$ in two ways. On one hand, since the circles touch internally, the distance between their centers is the difference of their radii, i.e., $O Q=R-r$. On the other hand, from the right trapezoid $C H Q O$, we get that $O Q^{2}=C H^{2}+(Q H-O C)^{2}=1+x^{2}$.

Thus, $\sqrt{1+x^{2}}+\sqrt{4 x^{2}+1}=\sqrt{x^{2}+16}$, from which $5 x^{2}+2+2 \sqrt{4 x^{4}+5 x^{2}+1}=x^{2}+16 \Leftrightarrow \sqrt{4 x^{4}+5 x^{2}+1}=7-2 x^{2}$.

Under the condition $7-2 x^{2} \geq 0$, the last equation is equivalent to the following: $4 x^{4}+5 x^{2}+1=49-28 x^{2}+4 x^{4}$, $x^{2}=\frac{16}{11}$. Then we get that $R=8 \sqrt{\frac{3}{11}}, r=5 \sqrt{\frac{3}{11}}$.