The base AC, which is a chord of the circumscribed circle, can be bisected and the perpendicular bisector will meet the apex B, because triangle ABC is isosceles and the bisector splits the triangle into two congruent right-angled triangles. The bisector of chord AC also passes through the centre of the circle, O. BO=AO=OC=r, radius of the circle. Angle AOC is twice ABC because the angle at the centre is twice the angle at the circumference for sector ABC. AB=BC=12m. Angle ABC is bisected by the side bisector on AC. If AC's bisector is at X on AC, then AX=XC=10m and sinABX=AX/AB=10/12=5/6. Angle AOX=twice angle ABX [or AOX=180-twice angle ABX], because angle AOC=2ABC and ABC=2ABX and AOC=2AOX. It follows that angle AOX=ABC=2sin^-1(5/6). AX/AO=10/r=sinAOX.
Trig identities: sin2y=2sinycosy and sin(y)=sin(180-y).
Applying the first of these identities: sinAOX=sinABC=sin2ABX=2sinABXcosABX=2*5/6*sqrt(1-(5/6)^2).
Therefore r=10/(2sinABXcosABX)10/(2*5/6*sqrt(1-(5/6)^2))=10.8544m approx. (36/sqrt(11) or 36sqrt(11)/11).
(In actual fact AOC as the angle at the centre is a reflex angle and ABC is an obtuse angle. This does not alter the logic because the sine of an angle is the same as the sine of its complement. BO, the radius, is longer than BX. Angle AOX=180-2ABX, so AX/AO=sin(180-2ABX)=sin(2ABX).)