$\widehat{AC}$ and $\widehat{BC}$ are arcs with centers $B$ and $A$ respectively. The circle in the figure passes through the midpoint of $AB$ and touches both the arcs. If $AB=12$, find the radius of the circle.
Let AO = x + r
OC = r
AC = 6
By the Pythagorean Theorem
r^2 + 6^2 = (r + x) ^2
r^2 + 36 = r^2 + 2rx + x^2
36 = 2rx + x^2
x^2 + 2rx - 36 = 0
x^2 + 2rx = 36 complete the square on x
x^2 + 2rx + r^2 = 36 +r^2
(x + r)^2 = 36 + r^2
x + r = sqrt [36 + r^2 ]
x = sqrt [36 + r^2 ] - r
x + 2r = 12
sqrt [ 36 + r^2 ] - r + 2r = 12
sqrt [ 36 +r^2] = 12 - r
36 + r^2 = r^2 -24r + 144
24r = 108
r =108 / 24 = 9/2