Two concentric circles have radii $13$ and $8$. $A$ chord of the larger circle touches the smaller circle. Find the length of the chord.

(N/A) Let $P$ be the common center of the two concentric circles. Let $\overline{AB}$ be the chord of the larger circle with radius $R = 13$,which touches the smaller circle with radius $r = 8$ at point $M$.
Since $\overline{AB}$ is a tangent to the smaller circle at $M$,the radius $\overline{PM}$ is perpendicular to the chord $\overline{AB}$. Thus,$\angle PMB = 90^{\circ}$.
In the right-angled triangle $\Delta PMB$:
By the Pythagorean theorem,$PB^2 = PM^2 + MB^2$.
Here,$PB = 13$ (radius of the larger circle) and $PM = 8$ (radius of the smaller circle).
$13^2 = 8^2 + MB^2$
$169 = 64 + MB^2$
$MB^2 = 169 - 64 = 105$
$MB = \sqrt{105}$.
Since the perpendicular from the center to a chord bisects the chord,$AB = 2 \times MB$.
$AB = 2\sqrt{105}$.