Since $AB$ is a diameter and both $AD$ and $BC$ are tangents, we have $AD \perp AB$ and $BC \perp AB$. Basically we have trapezoid $ABCD$ with two right angles, we have to find $AB$ using $AD$ and $BC$. Let $O = BD \cap AC$. Let $OA = y. OB = q, OC = x, OD = p$. Since AB is the diameter ,  $\therefore \angle AOB = 90^\circ$ . 

Because of the fact that $\triangle AOD$ and $\triangle BOC$ are right angled triangles we have:

$a^2+b^2 = p^2 + y^2 + x^2 + q^2 = AB^2 + CD^2$

$\Rightarrow AB^2 = a^2 +b^2 - CD^2$

Also we have 

$CD^2 = p^2 + x^2$

$AB^2 = a^2 +b^2 - p^2 - x^2$

Since DA and CB are tangents , by using power of point with respect to the circumcircle of $\triangle AOC$ we have:

\[b^2 = x(x+y)\]

\[a^2 = p(p+q)\]

combining with the previous equation we get 

$AB^2 = p(p+q) + x(x+y) - p^2 - x^2$

$\Rightarrow AB^2 = pq + xy$

We have $AD \parallel BC$ so,

$\frac{p}{q} = \frac{y}{x}$

$\Rightarrow pq = \frac{q^2y}{x}$

Combining with the previous equation we get,

$AB^2 =\frac{q^2y}{x} +xy $

$\Rightarrow AB^2 = \frac{y}{x}(q^2+x^2)$

$\Rightarrow AB^2 = \frac{y}{x}b^2$

$\Rightarrow AB^2 = \frac{a}{b}b^2$

$\Rightarrow AB^2 = ab$

So the answer is $AB = \sqrt{ab}$