4. Given a circle of radius $R$. At a distance of $2 R$ from the center of the circle, a point $A$ is chosen. From this point, a tangent and a secant are drawn, and the secant is equidistant from the center of the circle and the point of tangency. Find the length of the segment of the secant enclosed within the circle.
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_5455afdcd76b21a5003cg-2.jpg?height=651&width=899&top_left_y=1965&top_left_x=270)

Let $O$ be the center of the circle, $B$ the point of tangency, $CG$ the secant, $BF$ and $OD$ perpendicular to the secant, and $E$ the intersection point of the secant with the radius $OB$. Segments $OD$ and $BF$ are equal and parallel, so $ODBF$ is a parallelogram and $OE = BE = \frac{R}{2}$. Denoting the angle $DOE$ by $\alpha$, we get:
$OD = \frac{R}{2} \cos \alpha, CG = 2 CD = 2 \sqrt{OC^2 - OD^2} = R \sqrt{4 - \cos^2 \alpha}$.

The sides of angles $BAE$ and $DOE$ are pairwise perpendicular, so angle $BAE$ is also equal to $\alpha$. From the right triangle $BAE$ we find:

$\operatorname{tg} \alpha = \frac{BE}{AB} = \frac{BE}{\sqrt{OA^2 - OB^2}} = \frac{R}{2 \sqrt{4R^2 - R^2}} = \frac{1}{2 \sqrt{3}}$,

$\cos^2 \alpha = \frac{1}{1 + \operatorname{tg}^2 \alpha} = \frac{12}{13}, CG = R \sqrt{4 - \frac{12}{13}} = 2R \sqrt{\frac{10}{13}}$.

Answer: $2R \sqrt{\frac{10}{13}}$.

Remarks for checking. It is proven that $ODBF$ is a parallelogram and the equality of angles $BAE$ and $DOE$ - 2 points.