Two circles with radii 5 and 3 touch each other internally. A chord of the larger circle touches the smaller circle and is divided by the point of tangency in the ratio $3: 1$. Find the length of this chord.
Let $O_{1}$ and $O_{2}$ be the centers of circles with radii 5 and 3, respectively. $AB$ is a given chord, $C$ is the point of tangency with the smaller circle ($AC: BC=1: 3$), and $P$ is the projection of point $O_{1}$ onto the radius $O_{2}C$ of the smaller circle.

If $AC = x$, then $BC = 3x$. Drop a perpendicular $O_{1}K$ to the chord $AB$. Then $AK = KB = 2x$, $CK = x$, $O_{1}P = CK = x$, $O_{1}O_{2} = 2$, $O_{2}C = 3$, and $O_{1}K^2 = O_{1}B^2 - KB^2 = 25 - 4x^2$.

By the Pythagorean theorem, $O_{1}O_{2}^2 = O_{2}P^2 + O_{1}P^2$, or $4 = \left(3 - \sqrt{25 - 4x^2}\right)^2 + x^2$, from which $x^2 = 4$. Therefore, $AB = 4x = 8$.
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## Answer

8.