Question 1125549
{{{ drawing(400,400, -12, 12, -12, 12, grid(0),
                   line(-5,-9.2, 5,-9.2),
               line(-10,-3.2, 10,-3.2),
               locate(-11,-3.5, "A"),
              locate(-6,-9.5, "B"),
              locate(0.3,1, "O"),
              green(line(0,0, -10,-3.2)),
              green(line(0,0, -5,-9.2)),
              locate(0.5,-6.1, "d=6"),
              locate(-5,-3.4, "10"),
              locate(-2.5, -9.4, "5"),
              locate(0.4, -1.6, "k"),
              locate(-6.2, -0.8, "r"),
              locate(-4.0, -5.8, "r"),
              circle(0,0,10.515)

)
}}}

<br>
From the figure (I tried to get it close to true-to-scale):

   (1)       {{{ r^2 = 10^2 + k^2 }}}
   (2)       {{{ r^2 = 5^2 + (k+6)^2 }}}<br>

Setting these two equal and solving for k,   k = 13/4  —>  {{{ highlight( r = sqrt(1769)/4 ) }}}cm  (approx 10.515cm)<br>


Check:
        (1)   {{{  10^2 + (13/4)^2 = 100 + (169/16) = (1600/16) + (169/16) = 1769/16 =  r^2 }}}  (ok)
        (2)  {{{  5^2 + ((13/4) + 6)^2 = 25 + ((13+24)/4)^2 = (400/16) + (37/4)^2 = (400 + 1369)/16 = 1769/16 = r^2 }}}  (ok)

—————
Although it is possible for some parallel chord problems to have a 2nd solution where the chords are on opposite sides of the circle's center, this problem does not have solution for that configuration.  <br>