Question 200379
A  sketch  would  help,  but  perhaps  you  could  make  one as  we  go.
.
We  understand  that  two  circles  of  radius, r  & 2r  intersect.  You  want  to  define  the  length  of  the  common  chord  thru both  centers, k.
.
If  the  circles  just  touched,  the  distance  would  be  the  sum  of  the  diameters,
 (r+r) + ( 2r + 2r)=6r
.
However  as  they  overlap,  this  is  diminished  by  the  distance  from  the  circle at  tangency,  to  the  intersect  with  the  perpendicular  chord.  If  we  LET   that  be  "a"  for  the  small  circle  and  "b"  for  the  large  circle,, 
 k=  6r -a -b.
.
Now,  to  find  a & b.  Looking  at  the  small  circle  first,  THE  LINE is  perpendiculat  to  a  chord  of  length  2c.  The  1/2   chord  is  of  course  "c" .  The  triangle  created  is  familiar, one  leg  "c" ,
one  leg (hypotenuse) radius,  and  third leg  unknown,, LET it  be  "e".
.
Using  Pythagorous  Theorem,  e^2 =  R^2 - c^2,  or  e = sqrt(r^2 -c^2).
.
Going  back  to  the  sketch,  e + a  = r,  or  e= r-a but  also  = sqrt (r^2 - c^2),,,,
Therefore a = r - sqrt( r^2 -c^2)
.
Likewise  for  larger  circle (r=2R),,,,,,b= 2r - sqrt( (2r)^2 -c^2)
.
Going  back  to  k = 6r -a -b,,,  substitute  in  and  simplify  to  
.
k = 3R +sqrt( R^2 -c^2) + sqrt( 4R^2 -c^2 )
.
.
.
The  answer  can  now  be  seen  as  k= R + 2R + e + d  (  legs  of  triangles )  ok