Question 344990
{{{drawing(400,288,-11,39,-18,18,
locate(0,0,A), locate(21,0,B), locate(5.5,10,C), locate(5.5,-8.5,D),
circle(0,0,10), circle(21,0,17), line(0,0,21,0), line(6,8,6,-8),
locate(6.5,0,E) 


)}}}

First we will assume the smaller circle is the one with radius 10.
We will let the radius of the larger circle be r. We draw in a
radius of each circle (in green):
 
Then AC = 10, AB=21, BC = r, and CE = 8 since it is half of
the common chord CD, which is 16.

{{{drawing(400,288,-11,39,-18,18, locate(6.5,0,E),
locate(0,0,A), locate(21,0,B), locate(5.5,10,C), locate(5.5,-8.5,D),
circle(0,0,10), circle(21,0,17), line(0,0,21,0), line(6,8,6,-8),
green(line(0,0,6,8),line(6,8,21,0),locate(.5,5,10)), locate(14,6,r),
locate(6.5,4,8)

 )}}}

Triangle ACE is a right triangle, so we can find AE by the Pythagorean
theorem:

{{{AC^2 = AE^2+CE^2}}}

{{{10^2 = AE^2+8^2}}}
{{{100=AE^2+64}}}
{{{36=AE^2}}}
{{{sqrt(36)=AE}}}
{{{6=AE}}}

{{{drawing(400,288,-11,39,-18,18, locate(6.5,0,E),
locate(0,0,A), locate(21,0,B), locate(5.5,10,C), locate(5.5,-8.5,D),
circle(0,0,10), circle(21,0,17), line(0,0,21,0), line(6,8,6,-8),
green(line(0,0,6,8),line(6,8,21,0),locate(.5,5,10)), locate(14,6,r),
locate(6.5,4,8), locate(3,0,6)

 )}}}

<i>(We can see here that the larger circle could not have been the
one with radius 10, for that would have made BE = 6 which would have
been shorter than AE.)</i>

Therefore since AB = 21 and AE = 6

AB = AE + EB

21 = 6 + EB

15 = EB

{{{drawing(400,288,-11,39,-18,18, locate(6.5,0,E),
locate(0,0,A), locate(21,0,B), locate(5.5,10,C), locate(5.5,-8.5,D),
circle(0,0,10), circle(21,0,17), line(0,0,21,0), line(6,8,6,-8),
green(line(0,0,6,8),line(6,8,21,0),locate(.5,5,10)), locate(14,6,r),
locate(6.5,4,8), locate(3,0,6), locate(13,0,15)

 )}}}


Triangle CEB is a right triangle, so we can find r by
the Pythagorean theorem

{{{BC^2 = BE^2+CE^2}}}
{{{r^2 = 15^2+8^2}}}
{{{r^2=225+64}}}
{{{r^2=289}}}
{{{r=sqrt(289)}}}
{{{r=17cm}}}

Edwin</pre>