Draw in altitude AD which will pass through center O,
and bisect the base into 2 segments each  units in length,
since the triangle is isoceles.  So segment BD is  units.
We will also draw in radius OB, and label it r.  Radius
OA also has length r. We will label the length of OD as h. 





ADB and ODB are both right triangles, so using the
Pythagorean theorem

BD² + AD² = AB² and BD² + OD² = OB²

Since AD = OA + OD, we have:

BD² + (OA + OD)² = AB² and BD² + OD² = OB²

In terms of the lengths of the sides we have this
system of equations to solve:



Simplify the first equation:



Multiply through by LCD=4



Simplify the second equation:



Multiply through by LCD=4


So now our system to solve becomes:



We solve the second equation of the system for 4h²:




And substitute that for 4h² in the first equation of the system:



and simplify:


Divide through by 4


Now we solve

 for 2h by the principle of square roots,
(we only take positive square roots:



Substitute for 2h in




Square both sides:


Simplify:





, with denominator rationalized, if you like, as


or approximately:
r=6.623632224

Edwin