External Tangents

Let $O_1$ be the center of the green circle.

Let $O_2$ be the center of the orange circle.

Let the point of intersection of line $O_1O_2$ and $CD$ be $P$ .

Note that the line $O_1O_2$ is the same length as $AB$ since $AB$ is a common tangent.

$\bigtriangleup$ $CO_1P$ is congruent to $\bigtriangleup$ $PDO_2$ since $CO_1$ and $DO_2$ are both length 5, $\angle$ $DCO_1$ = $\angle$ $CDO_2$ = $90$ , and $\angle$ $CPO_1$ = $\angle$ $DPO_2$ since they are opposite angles.

$\therefore$ $CP$ = $PD$ = $8$

We know that $CO_1$ = $5$

So $O_1P^2$ = $8^2+5^2$ = $89$

$AB^2$ = $(2 \cdot O_1P)^2$ = $4 \cdot O_1P^2$ = $4 \cdot 89$ = $\boxed{356}$

Another way is to create an equation from two similar kites. From the similar kites we find $25=(16+x)x$ . The length $AB= 16+2x$ , so $AB^2 = (16+2x)^2=16^2+4(16+x)x=16^2+4(25)=356$