A line intersects the circles $k_{1}$ and $k_{2}$, which have no common points, at equal-length chords $A B$ and $C D$, respectively. The intersection points are in order $A, B, C$, and $D$. For a point $P$, it holds that $P A$ is tangent to $k_{1}$ and $P D$ is tangent to $k_{2}$. Determine the ratio $\frac{P A}{P D}$ using the radii of the two circles.
Solution. The points $A, D$, and $P$ form a triangle; otherwise, the line $AD$ would be a common tangent to the two circles, and in that case, the chords $AB$ and $CD$ would degenerate into points.

Using the notation from Figure 1, let $\angle PAD = \alpha$ and $\angle ADP = \beta$. Points $E$ and $F$ are points on circles $k_1$ and $k_2$, respectively, in the half-plane bounded by line $AD$ that does not contain point $P$.

![](https://cdn.mathpix.com/cropped/2024_05_02_0c65395a900d4c244de2g-1.jpg?height=759&width=892&top_left_y=325&top_left_x=606)

Figure 1

Then, by the theorem of the angles subtended by a tangent and a chord, $\angle AEB = \alpha$ and $\angle CFD = \beta$. Let $AB = CD = x$, and the radii of the circles be $r_1$ and $r_2$, respectively. Thus,

$$
\sin \alpha = \frac{x}{2 r_1} \quad \text{and} \quad \sin \beta = \frac{x}{2 r_2}
$$

Applying the Law of Sines in triangle $APD$:

$$
\frac{PA}{PD} = \frac{\sin \beta}{\sin \alpha} = \frac{\frac{x}{2 r_2}}{\frac{x}{2 r_1}} = \frac{r_1}{r_2}
$$

Therefore, the ratio of the lengths of the tangent segments is equal to the ratio of the radii of the circles.

Remark. Many made the mistake of not considering all possible placements, for example, they did not address the placement shown in Figure 2. Thus, they often wrote down relationships for the angles that do not hold in this case (although with minor adjustments, they would be true). Therefore, a point was deducted for this error.

![](https://cdn.mathpix.com/cropped/2024_05_02_0c65395a900d4c244de2g-1.jpg?height=450&width=905&top_left_y=1756&top_left_x=599)

Figure 2