This is easy to solve. . Extend the intersection circle of AB and D. Then BD is the diameter, and connecting CD gives the angle DCB = 90.

From the tangent OC⊥AB, angle OCA = 90 = angle DCB.

* * * Subtract angle OCB to get angle DCO= angle BCA. And DO = CO (radius), and the angle D= angle BCA is obtained. So triangle ACB∽ triangle ADC (two angles), so AB*AB=AC square (derived from AD/AC=AC/AB).

The arc length of BC is 20/9π, and the angle COB=50 is obtained.

AC=AO*sin50

List the equation AB * (AB+ 16) = (AB+8) SIN 50 square. The number of quadratic equations is not easy to calculate.