A hexagon with sides of length 2, 7, 2, 11, 7, 11 is inscribed in a circle. Find the radius of the circle.

*As suggested, *if* it matters, you may assume that the sides listed are given in order*

The sides {7, 2, 11} may be inscribed in a semicircle of the same radius as the original circle.

Let AB = d be the diameter of the semicircle, BC = 11, CD = 2, and AD = 7. Draw AC of length x, and BD of length y.

Note that angles ADB and ACB are right angles. Hence, by Pythagoras:

x^{2} = d^{2} - 121

y^{2} = d^{2} - 49

Ptolemy's Theorem states that in a cyclic quadrilateral the sum of the products of the two pairs of opposite sides equals the product of its two diagonals.

Hence 2d + 77 = xy.

Therefore (2d + 77)^{2} = (d^{2} - 121)(d^{2} - 49).

Simplifying, and dividing by d, we obtain d^{3} - 174d - 308 = 0.

This factorizes, giving (d - 14)(d^{2} + 14d + 22) = 0.

The quadratic factor has negative real roots, so d = 14 is the only positive real root.

Therefore the radius of the circle in which the hexagon is inscribed is 7.