8,9 On the coordinate plane ( $x ; y$ ), a circle of radius 4 with its center at the origin is drawn. A line given by the equation $y=4-(2-\sqrt{3}) x$ intersects it at points $A$ and $B$. Find the sum of the lengths of the segment $A B$ and the smaller arc $A B$. #
Solving the system of equations, find the coordinates of the points of intersection of the line and the circle. Then, applying the cosine theorem, find the angle between the radii of the circle drawn to the found points of intersection.

## Solution

Solving the system of equations

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=16 \\
y=4-(2-\sqrt{3}) x
\end{array}\right.
$$

we find the coordinates of the points of intersection of the line and the circle: \( A(0 ; 4), B(2 ; 2 \sqrt{3}) \). Then \( AB = \sqrt{(0-2)^{2}+(4-2 \sqrt{3})^{2}} = \sqrt{32-16 \sqrt{3}} = 4 \sqrt{2-\sqrt{3}} \). Let \( O \) be the origin. By the cosine theorem in triangle \( AOB \), we find that \( \cos \angle AOB = \frac{OA^{2} + OB^{2} - AB^{2}}{2 \cdot OA \cdot OB} = \frac{16 + 16 - 32 + 16 \sqrt{3}}{2 \cdot 4 \cdot 4} = \frac{\sqrt{3}}{2} \). Therefore, the degree measure of the smaller arc \( AB \) is \( 30^{\circ} \). The length of this arc is one twelfth of the circumference of a circle with radius 4, i.e., \( \frac{2 \pi}{3} \). Thus, the required sum is \( \frac{2 \pi}{3} + 4 \sqrt{2-\sqrt{3}} \).

![](https://cdn.mathpix.com/cropped/2024_05_06_b41405b02b6e888d68cag-08.jpg?height=692&width=706&top_left_y=0&top_left_x=670)

## Answer

\(\ \mid f r a c\{2 \backslash$ pi $\}\{3\}+4 \mid\) sqrt \{2-\mid \operatorname{sqrt}\{3\}\} \$$ --> \(\frac{2 \pi}{3} + 4 \sqrt{2-\sqrt{3}}\).