I think the solution lies more in a diagrammatic representation than anything else. From the distances, one thing is evident - the villages are non-linear. So A, B and C form the three vertices of a triangle (It would have been better if A, B and C had been houses rather than villages since it is difficult to imagine villages as vertices of a triangle).

Anyway, the point equidistant from the three vertices of a triangle is the circumcentre. The circumcentre is found by drawing perpendiculars from the midpoints of each segment of the triangle. The circumcentre is the point where the three perpendiculars intersect. The property of the circumcentre is that a circle can be drawn such that all the three vertices lie on the circumference. Therefore, the circumradius is the length which is equidistant from the three vertices.

If all the three lengths of a triangle are known, then the circumradius is

(abc)/sqrt [(a+b+c) (b+c-a) (c+a-b) (a+b-c). The formula is mentioned by.

Here a, b and c are the lengths of the sides of the triangle. Therefore a = 8, b = 7 and c = 5.

Thus the circumradius is (8*7*5)/Sqrt [(8+7+5) (7+5-8) (5+8-7) (8+7-5)

this is equal to 280/sqrt(20*4*6*10)

= 280/sqrt (4800)

= 280/69.28

= 4.04