Question 1189421
<pre>

Let P be the center of the circle.  Here are the 
given parts drawn to scale: 

{{{drawing(400,400,-22,22,-22,22,
locate(-20,.7,A),locate(19.2,.7,B),locate(-3.9,20.4,D),
locate(-10.9,0,O),locate(-16,-11,C),circle(0,0,.4),locate(0,-.5,P),
locate(-7.2,9.5,20),locate(-13.2,-5.6,12),locate(-15,1.7,8),

circle(0,0,19), line(-19,0,19,0), line(-41/11,18.63081957,-169/11,-11.17849174)
)}}}

By the intersecting chord theorem,

{{{AO*OB}}}{{{""=""}}}{{{CO*OD}}}
{{{8*OB}}}{{{""=""}}}{{{12*20}}}
{{{8*OB}}}{{{""=""}}}{{{240}}}
{{{OB}}}{{{""=""}}}{{{30}}}

{{{AB}}}{{{""=""}}}{{{AO+OB}}}{{{""=""}}}{{{8+30}}}{{{""=""}}}{{{38}}}{{{""=""}}}{{{diameter}}}

So radius = half the diameter = 19 cm, AP = 19 = BP

{{{OP}}}{{{""=""}}}{{{AP-AO}}}{{{""=""}}}{{{19-8}}}{{{""=""}}}{{{11}}}

We put those values in the drawing and draw radius CP (in green).
CP is a radius so CP = 19 cm.

{{{drawing(400,400,-22,22,-22,22,
locate(-20,.7,A),locate(19.2,.7,B),locate(-3.9,20.4,D),
locate(-10.9,0,O),locate(-16,-11,C),circle(0,0,.4),locate(0,-.5,P),
locate(-7.2,9.5,20),locate(-12.9,-4.2,12),locate(-15,1.7,8),
green(line(0,0,-169/11,-11.17849174)), locate(-5.75,1.7,11),locate(-8.5,-5.5,19),
circle(0,0,19), line(-19,0,19,0), line(-41/11,18.63081957,-169/11,-11.17849174)
)}}}

We find the central angle APC of arc AC by using the law of 
cosines on &Delta;COP, the case is SSS:

{{{cos("<APC")}}}{{{""=""}}}{{{(OP^2+CP^2-OC^2)/(2*OP*CP)}}}

{{{cos("<APC")}}}{{{""=""}}}{{{(11^2+19^2-12^2)/(2*11*19)}}}

{{{cos("<APC")}}}{{{""=""}}}{{{338/418}}}

{{{"<APC"}}}{{{""=""}}}{{{36.03941623^o}}}{{{""=""}}}
{{{matrix(1,2,0.6290064738,radians)}}} 

The formula for the arc length is {{{s}}}{{{""=""}}}{{{r*theta}}}
where &theta; is in radians.  So

{{{matrix(1,4,length,of,arc,AC)}}}{{{""=""}}}{{{19*0.6290064738}}}

{{{matrix(1,4,length,of,arc,AC)}}}{{{""=""}}}{{{matrix(1,2,11.951123,cm)}}}

Edwin</pre>