7. The distance from the point of intersection of the diameter of a circle with radius $11 \mathrm{~cm}$ and a chord of length 18 cm to the center of the circle is 7 cm. In what ratio does the point of intersection divide the chord?
Solution: $R=11, AB=18, PO=7, \frac{AP}{PB}=$ ?

$AP \cdot PB=DP \cdot PC, DP=DO-PO=11-7=4, PC=11+7=18$,

$\left\{\begin{array}{l}AP \cdot PB=4 \cdot 18 \\ AP+PB=18\end{array}, AP(18-AP)=72,18AP-AP^{2}=72\right.$,

![](https://cdn.mathpix.com/cropped/2024_05_06_8eeeb8d6ace591217257g-13.jpg?height=317&width=374&top_left_y=752&top_left_x=1569)

$AP^{2}-18AP+72=0, AP=\frac{18 \pm \sqrt{18^{2}-4 \cdot 4 \cdot 18}}{2}=\frac{18 \pm 6}{2}=\left[\begin{array}{l}12 \\ 6\end{array}, \frac{12}{6}=2 \text{ or } \frac{6}{12}=\frac{1}{2}\right.$.

Answer: $2:1, 1:2$.