$P$ lies in the exterior of $\odot(O, 8)$ such that $OP = 17$. Two tangents are drawn to the circle which touch the circle at $A$ and $B$. Find $AB$.

(240/17) Let $R$ be the intersection of $OP$ and $AB$. Since $OP$ is the perpendicular bisector of the chord $AB$,$OP \perp AB$ at $R$.
In right-angled $\Delta OAP$,$\angle OAP = 90^{\circ}$.
By Pythagoras theorem,$OP^2 = OA^2 + AP^2$.
$(17)^2 = (8)^2 + AP^2 \implies 289 = 64 + AP^2 \implies AP^2 = 225 \implies AP = 15$.
In $\Delta OAP$,$AR$ is the altitude to the hypotenuse $OP$.
Using the property of right triangles,$OA^2 = OR \cdot OP$.
$(8)^2 = OR \cdot 17 \implies OR = \frac{64}{17}$.
In right-angled $\Delta OAR$,$AR^2 = OA^2 - OR^2$.
$AR^2 = 8^2 - \left(\frac{64}{17}\right)^2 = 64 - \frac{4096}{289} = \frac{18496 - 4096}{289} = \frac{14400}{289}$.
$AR = \sqrt{\frac{14400}{289}} = \frac{120}{17}$.
Since $OP$ bisects $AB$,$AB = 2 \cdot AR = 2 \cdot \left(\frac{120}{17}\right) = \frac{240}{17}$.