4. In triangle $ABC$, side $AB$ is equal to $\sqrt{11}$, and the angle at vertex $A$ is twice the angle at vertex $C$. A tangent line $\ell$ to the circumcircle $\Omega$ of triangle $ABC$ is drawn through vertex $B$. The distances from points $A$ and $C$ to this tangent line are in the ratio $9: 25$.

a) Find the ratio of the distances from point $A$ to the lines $\ell$ and $BC$.

b) Find the distance from point $C$ to the line $\ell$ and the radius of the circle $\Omega$.
Answer. a) $9: 16$, b) $d_{C}=\frac{275}{54}, R=3$.

Solution. Let $\angle A C B=\gamma$, then $\angle B A C=2 \gamma$. By the generalized sine theorem, we find that $A B=2 R \sin \gamma, B C=2 R \sin 2 \gamma$. Note that $\gamma<90^{\circ}$ (otherwise the sum of the angles in the triangle would be greater than $180^{\circ}$).

By the theorem on the angle between a tangent and a chord, the angle between the tangent and $A B$ is $\gamma$, and the angle between the tangent and $B C$ is $2 \gamma$. Therefore, the distance $d_{A}$ from point $A$ to the tangent is $A B \sin \gamma = 2 R \sin ^{2} \gamma$. Similarly, the distance $d_{C}$ from point $C$ to the tangent is $B C \sin 2 \gamma = 2 R \sin ^{2} 2 \gamma$.

Thus, $\frac{25}{9}=\frac{d_{C}}{d_{A}}=4 \cos ^{2} \gamma$, from which $\cos ^{2} \gamma=\frac{25}{36}$. Then $\sin ^{2} \gamma=\frac{11}{36}, R=\frac{A B}{\sin 2 \gamma}=3$. Therefore, $d_{A}=\frac{11}{6}, d_{C}=\frac{275}{54}$.

The distance from point $A$ to line $B C$ is

$$
A B \cdot \sin \angle A B C=A B \sin 3 \gamma=A B\left(3 \sin \gamma-4 \sin ^{3} \gamma\right)=\frac{88}{27}
$$

The ratio of the distances from point $A$ to lines $\ell$ and $B C$ is $\frac{11}{6}: \frac{88}{27}=\frac{9}{16}$.