3 [ Properties and characteristics of the tangent ]

In a circle with a radius of 5, a chord $AB=8$ is drawn. Point $C$ lies on the chord $AB$ such that $AC: BC=1: 2$. Find the radius of the circle that is tangent to the given circle and touches the chord $AB$ at point $C$.
Let $E$ be the projection of the center $O$ of a given circle onto the chord $AB$. Then $E$ is the midpoint of $AB$ and $OE = \sqrt{OB^2 - BE^2} = \sqrt{25 - 16} = 3$.

If the desired circle with center $Q$ and radius $r$ touches the given circle at point $D$, then

$$
OQ = OD - QD = 5 - r, \quad CE = AE - AC = 4 - \frac{1}{3} \cdot 8 = \frac{t}{3}
$$

Let $F$ be the projection of point $O$ onto the line $QC$. Then $OFC E$ is a rectangle, so $CF = OE = 3$ and $OF = CE = \frac{\frac{1}{3}}{3}$. Consider the case when points $O$ and $Q$ lie on opposite sides of the line $AB$ (Fig.1). Then $QF = QC + CF = QC + OE = r + 3$. By the Pythagorean theorem, $OQ^2 = QF^2 + OF^2$, or $(5 - r)^2 = (r + 3)^2 + \frac{16}{9}$. From this equation, we find that $r = \frac{8}{9}$.

If points $O$ and $Q$ lie on the same side of the line $AB$ (Fig.2), then similarly we obtain the equation $(5 - r)^2 = (r - 3)^2 + \frac{16}{9}$, from which we find that $x = \frac{32}{9}$.

## Answer

$\frac{8}{9}$ or $\frac{32}{9}$.