A circle with diameter  $PQ$ of length 10 is internally tangent at P to a circle of  radius 20. Square ABCD is constructed with A and B on the large circle. CD is tangent at $Q$  to the smaller circle, and the smaller circle is outside ABCD. The length of AB can be written in the form of $m+\sqrt{n},$  where $m$ and  $n$ are integers, the m + n is

 
Call the center of the larger circle O. extend the diameter $\overline{pQ}$  to the other side of the square
(at point E), and draw  $\overline{AO}$. We now have right triangle, with hypotenuse of length 20.
Since   $OQ=OP-PQ=20-10=10,$
We know that  $OE=AB-OQ=AB-10$
The other leg. AE , is just $\frac{1}{2}$ AB 
Apply the Pythagorean theorem:
${(AB-10)}^2+{\left(\frac{1}{2}AB\right)}^2={20}^2$ 
 $A{B}^2-20AB+100+\frac{1}{4}A{B}^2-400=0$
 $A{B}^2-16AB-240=0$
The quadratic formula shows that the answer is 
 $\frac{16\pm \sqrt{{16}^2+4.240}}{2}=8\pm \sqrt{304}.$