Geometry problem with two chords originating from a common point outside the circle. This one's got me pulling my hair.
By the power-of-a-point theorem,

$$OA\cdot OB=OP\cdot OQ.$$
Construction: Join BQ and AP

Proof

$1. \angle BQO=\angle PAO$(Angle subtended betwewn a chord and a tangent is equal to the angle subtended by the chord on its alternate segment),

$2.\angle QBO=\angle OPA$(Angle subtended betwewn a chord and a tangent is equal to the angle subtended by the chord on its alternate segment)

SO THEY ARE SIMILAR

$\frac{AO}{OQ}=\frac{OP}{OB}$

$\frac {AO}{12}=\frac {4}{18}$

Solve it to get value of AO.