Question 194449
<pre><font size = 4 color = "indigo"><b>
Let the circles' centers be A and B.
Draw the common chord CD. Draw radii AC and BC.

We know that {{{AC=13}}}, {{{BC=15}}}, {{{AB=14}}} 

{{{drawing(400,400,-14,30,-22,22, 
line(0,0,14,0), locate(0,0,A), locate(14,0,B), 
circle(0,0,13), circle(14,0,15), line(5,12,5,-12),
line(5,12,0,0),line(5,12,14,0), locate(5,14,C),
locate(5,-12,D), locate(5.3,-.1,E) )}}}

We are given all three sides of triangle ABC, so
we can use the law of cosines to find cos(A)

{{{BC^2=AB^2+AC^2-2*AB*AC*cos(A)}}}

{{{2*AB*AC*cos(A)=AB^2+AC^2-BC^2}}}

{{{cos(A)=(AB^2+AC^2-BC^2)/(2*AB*AC)}}}

{{{cos(A)=(14^2+13^2-15^2)/(2*14*13)}}}

{{{cos(A)=(196+169-225)/(2*14*13)}}}

{{{cos(A)=140/364}}}

{{{cos(A)=5/13}}}

We know that the line through their centers
is the perpendicular bisector of the common 
chord CD.  

This tells us two things. 

1. If we find CE we can just
double it to get the common chord CD.  

2. Triangle AEC is a right triangle and therefore

{{{(CE)/(AC)=sin(A)}}}

{{{CE=AC*sin(A)}}}

{{{CE=13*sin(A)}}}

Now we use the identity 

{{{Sin^2(A) + Cos^2(A) = 1}}}

{{{Sin^2(A) = 1 - Cos^2(A)}}}

Substitute {{{5/13}}} for {{{Cos(A)}}}

{{{Sin^2(A) = 1 - (5/13)^2}}}

{{{Sin^2(A) = 1 - 25/169}}}

{{{Sin^2(A) = 169/169-25/169}}}

{{{Sin^2(A) = 144/169}}}

{{{sin(A) = sqrt(144/169)}}}

{{{sin(A) = 12/13}}}

Now we substitute that in

{{{CE=13*sin(A)}}}

{{{CE=13*(12/13)}}}

{{{CE=cross(13)*(12/cross(13))}}}

{{{CE=12}}}

CE is half the common chord CD, so

the common chord CD is twice CE or 

2x12 or 24.

Edwin</pre>