The circle $k_{1}$ touches the circle $k_{2}$ internally at point $P$. The chord $AB$ of the larger circle touches the smaller circle at point $C$, and $AP$ and $BP$ intersect the smaller circle at points $D$ and $E$, respectively. We know that $AB=84$, $PD=11$, and $PE=10$. What is the length of $AC$?
I. solution. Since the tangents of the two circles at $P$ are common, $\angle PDE = \angle PAB$, because in the corresponding circles, these are the inscribed angles corresponding to the same arc, to which the inscribed angle formed by the tangent at $P$ also belongs (figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_5a1c3fa0bee9d1d74b36g-1.jpg?height=612&width=593&top_left_y=255&top_left_x=755)

Due to the similarity of triangles $ABP$ and $DEP$, we have

$$
PA : PB = PD : PE = 11 : 10
$$

which means

$$
PA = \frac{11}{10} PB \quad \text{and} \quad PD = \frac{11}{10} PE
$$

Therefore, by the secant theorem,

$$
\frac{AC^2}{BC^2} = \frac{AP \cdot AD}{BP \cdot BE} = \frac{PA(PA - PD)}{PB(PB - PE)} = \frac{\frac{11}{10} PB \cdot \left(\frac{11}{10} PB - \frac{11}{10} PE\right)}{PB(PB - PE)} = \frac{11^2}{10^2}
$$

so $BC = \frac{10}{11} AC$. From this, $84 = AB = AC + BC = \frac{21}{11} AC$, hence $AC = \frac{11}{21} \cdot 84 = 44$.

II. solution. Apply the dilation centered at $P$ that maps the smaller circle to the larger one. In this process, the image of $D$ is $A$, and the image of $E$ is $B$, so

$$
PA : PB = PD : PE = 11 : 10
$$

The image of line $AB$ is parallel to $AB$ and is tangent to the larger circle; let the point of tangency be $F$. Point $F$ bisects the arc $AB$, because the tangent is perpendicular to the radius $OF$ at the point of tangency, which, being perpendicular to the tangent parallel to $AB$, lies on the perpendicular bisector of segment $AB$. Thus, $PF$ bisects angle $APB$. Since $AB$ is tangent to the smaller circle at $C$, the image of $C$ under the dilation is $F$, so $C$ lies on the angle bisector. By the angle bisector theorem, $AC : CB = PA : PB = 11 : 10$, so

$$
AC = \frac{11}{21} AB = \frac{11}{21} \cdot 84 = 44
$$

Juhász Máté Lehel (Budapest, Fazekas M. Főv. Gyak. Gimn., 11th grade)