5. Segment $A B=8$ intersects plane $\alpha$ at an angle of $30^{\circ}$ and is divided by this plane in the ratio $1: 3$. Find the radius of the sphere passing through points $A$ and $B$ and intersecting plane $\alpha$ in a circle of the smallest radius.
Answer: $2 \sqrt{7}$. Solution. Denoting the intersection point of $A B$ with the plane $\alpha$ as $C$, we get $A C=2, B C=6$. At the intersection of the sphere with the plane, we obtain a certain circle. Draw the diameter $M N$ of this circle through $C$. Then $A B$ and $M N$ are chords of the sphere, and by the property of intersecting chords: $M C \cdot C N=A C \cdot C B=12$. Since

$M N=M C+C N \geq 2 \sqrt{M C \cdot C N}=4 \sqrt{3}$, the minimum radius of the circle is greater than or equal to $2 \sqrt{3}$, and the value $2 \sqrt{3}$ is achieved when $M C=C N=2 \sqrt{3}$, that is, $C$ is the center of this circle.

Since $\angle C O P=90^{\circ}-\angle O C P=\angle N C P=30^{\circ}$, then

![](https://cdn.mathpix.com/cropped/2024_05_06_0f7f3ab80da5a0ce1d2dg-2.jpg?height=503&width=551&top_left_y=2164&top_left_x=1392)
$O C=2 \cdot C P$. Here, $C P=\frac{A B}{2}-A C=2$. Therefore, $R^{2}=O M^{2}=M C^{2}+O C^{2}=12+4^{2}=28$.

Answer for variant 152: $2 \sqrt{19}$.

Answer for variant 153: $2 \sqrt{13}$.

Answer for variant 154: $\sqrt{39}$.