4. Two tangents are drawn from a point $A$ outside a circle with a diameter of $10 \mathrm{~cm}$. The point $A$ is $1.3 \mathrm{~dm}$ away from the center of the circle. Calculate the distance between the points of tangency on the circle and express it in centimeters. Round the result to one decimal place.
4. We draw a sketch: the center of the circle $S$, a point $A$ outside the circle, and the points of tangency $B$ and $C$. We denote the angle $\angle B S A$ as $\alpha$, the distance from $S$ to $A$ as $x$, and the distance from $B$ to $C$ as $2 y$. Since triangle $A S B$ is a right triangle, we use the trigonometric function $\cos \alpha=\frac{r}{x}$, from which we calculate $\alpha=67.38^{\circ}$. Triangle $D S B$ is also a right triangle ($D$ is the intersection of segments $S A$ and $B C$). Using the trigonometric function $\sin \alpha=\frac{y}{r}$, we calculate $y=r \cdot \sin \alpha=4.6$.

![](https://cdn.mathpix.com/cropped/2024_06_07_59669f975316a35884bfg-09.jpg?height=374&width=596&top_left_y=267&top_left_x=1184)
We then calculate the distance $2 y=9.2$.

We can solve the problem differently: the area of the right triangle $A S C$ is equal to $\frac{1}{2}|C S| \cdot|C A|=\frac{1}{2}|A S| \cdot|C D|$, from which we express $|B C|=2|C D|=\frac{2 \cdot|S C| \cdot \sqrt{|S A|^{2}-|S C|^{2}}}{|S A|}=$ $\frac{2 r \sqrt{x^{2}-r^{2}}}{x}=\frac{2 \cdot 5 \cdot 12}{13} \approx 9.2$.

Drawn sketch .1 point

Expression $\cos \alpha=\frac{r}{x}=\frac{5}{13}$ .1 point

Calculation of the angle $\alpha=67.38^{0}$ 1 point

Expression $\sin \alpha=\frac{y}{r}$ 1 point

Calculation $y=4.6$ 1 point

Calculated distance $2 y=9.2$. 1 point