Question 388780
Consider the isosceles triangle ABC, with A the vertex angle.Then AB = AC = 13. If we let D be the midpoint of segment BC, then segment AD is the altitude of triangle ABC from A to segment BC, and AD = 5, as given.  This altitude AD must lie on a diameter, due to symmetry.  Let AE be the diameter that contains the altitude AD.  Then triangle ACE (in that order) is a right triangle, with C as the right angle.By similarity of the right triangle ACE with  the right triangle ADC,  we must have, by proportionality of corresponding sides, that {{{d/13 = 13/5}}}, or d = 169/5, or d = 33.8, where d = diameter of the circle.  Thus r = 16.9.
{{{drawing(400,400,-1.5,1.5, -1.5,1.5, 

circle(0,0,1), triangle(0,1,.67010309,.742268,-.67010309,.742268),
locate(.4,1,13), locate(-.4,1,13), locate(0,.9,5),
locate(.7,.9,C),locate(-.7,.9,B),line(0,-1,0,1),
line(0,-1,.67010309,.742268), locate(0,1.1,A), locate(0,-1.1,E), locate(-.1,.7,D), locate(0,.3,d)

 )}}}
Hope this helps!