Question 720934
<pre>
Let the common center of the two circles be O

{{{drawing(400,400,-14,14,-14,14,

circle(0,0,8),circle(0,0,13), locate(-13.8,.5,A), line(-13,0,13,0),
locate(13.5,.5,B), line(4.923076923,6.305815856,13,0),circle(0,0,.2),
line(4.923076923,6.305815856,-13,0),locate(4.8,7.6,D),locate(0,-.2,O),
locate(4,0,8),locate(10,0,5),locate(-4,0,8),locate(-10,0,5)
  )}}}

Draw OD.  Then  OD&#8869;BD because radius OD drawn to the point of tangency is
perpendicular to the tangent line BD.

so &#2017;OBD is a right triangle with hypotenuse OB.

cos(&#8736;DOB) = {{{OD/OB}}} = {{{8/13}}}

cos(&#8736;AOD) = cos(180°-&#8736;DOB) = -cos(&#8736;DOB) = {{{-8/13}}} 

{{{drawing(400,400,-14,14,-14,14,
green(line(0,0,4.923076923,6.305815856)),locate(2,4,8),
circle(0,0,8),circle(0,0,13), locate(-13.8,.5,A), line(-13,0,13,0),
locate(13.5,.5,B), line(4.923076923,6.305815856,13,0),circle(0,0,.2),
locate(4,0,8),locate(10,0,5),locate(-4,0,8),locate(-10,0,5),
line(4.923076923,6.305815856,-13,0),locate(4.8,7.6,D),locate(0,-.2,O)  )}}}

Use the law of cosines:

ADČ = AOČ + ODČ - 2·AB·OD·cos(&#8736;AOD)

ADČ = 13Č + 8Č - 2·13·8·{{{(-8/13)}}}

ADČ = 169 + 64 + 128

ADČ = 361

AD = &#8730;<span style="text-decoration: overline">361</span>

AD = 19.

Edwin</pre>