A Weird Cut Of Concentric Circles

Since $4\sqrt{13} > 8$ , the smaller circle has a radius 8 and the larger circle has a radius of $4\sqrt{13}$ .

Clearly, $BD$ is tangent to the smaller circle at $D$ . Therefore angle $BDO = 90^\circ$ . Then by Pythagorean theorem

$BD^2 + OD^2 = OB^2 \Rightarrow BD = \sqrt{OB^2 -OD^2} = \sqrt{(4\sqrt{13})^2 - 8^2} = \sqrt{208 - 64} = \sqrt{144} = 12$

Now extend $BD$ to a point E on the larger circle. Join $AD$ .

Since $OB$ was extended to meet the larger circle at $A$ , $AB$ is the diameter of the larger circle.

Since angle $AEB$ is enclosed in the semicircle, angle $AEB$ is $90 ^\circ$ .

In triangle $ODB$ and triangle $AEB$ , angle $OBD$ is common and angle $ODB$ and angle $AEB$ are both $90^\circ$ . Therefore triangle $ODB$ and triangle $AEB$ are similar. Then,

$\dfrac{OB}{AB} = \dfrac{OD}{AE} = \dfrac{BD}{BE} \Rightarrow \dfrac{4\sqrt{13}}{8\sqrt{13}} = \dfrac8{AE} =\dfrac{12}{BE}$

Clearly, $AE=16$ and $BE=24$ . $ED = BE-BD = 12.$

Triangle $AED$ is right angled at $E$ . Therefore

$AD = \sqrt{AE^2 + ED^2} = \sqrt{16^2 + 12^2} = \boxed{20} .$

The simplest method is by the use of coordinate geometry... take the circles with center as the origin then find sin theta and cos theta by trigonometry now the coordinates of A(12,8) by parametric form and B will be (0,-8) using distance formula find AB.

Assume as given in the sketch (but not in the description,) that angle ODB = 90 degrees.
$In ~rt. ~\Delta OBD ~~BD^2=(4\sqrt{13})^2 - 8^2=16*9 ~~\therefore ~CosOBD=\dfrac {\sqrt{16*9}}{4\sqrt{13}}=\dfrac 3 {\sqrt{13}}\\ Applying ~Cos ~Rule ~to ~\Delta ABD,\\ AD^2 = AB^2+BD^2 - 2*AB*BD*CosOBD\\ =4*16*13+16*9 - 2*(8\sqrt{13})*(4*3)*(\dfrac 3 {\sqrt{13}})=400.\\ \text{Smallest distance between two points is a st. line. } \therefore ~AD=20.$

Using the Pythagorean Theorem, BD = sqrt((4 sqrt(13)^2 - 8^2)) = 12. Angle B = arctan(8/12) = 33.69006753. Finally, using the law of cosines, (AD)^2 = (8 sqrt(13)^2 + 12^2 - 2 12 8 sqrt(13) cos(33.69006753) = 400, so AD = 20. Ed Gray