10. Point $C$ divides the diameter $A B$ in the ratio $A C: B C=2: 1$. A point $P$ is chosen on the circle. Determine the values that the ratio $\operatorname{tg} \angle P A C: \operatorname{tg} \angle A P C$ can take. In your answer, specify the smallest such value.
Answer: $\frac{1}{2}$.

Solution: The angle $\angle A P B=90^{\circ}$, as it subtends the diameter. Drop a perpendicular $C H$ to $A P$, it will be parallel to $P B$. Therefore, $\triangle A P B \propto \triangle A H C$ with a ratio of $\frac{2}{3}$. Hence, $\operatorname{tg} \angle A P C=$ $\frac{H C}{P C}=2 \frac{P B}{A P}=2 \operatorname{tg} \angle P A C$