
The circumradius of a cyclic quadrilateral with consecutive sides a, b, c, d is given by
R = (1/4)sqrt((ac+bd)(ad+bc)(ab+cd)/((sa)(sb)(sc)(sd))), where s=(a+b+c+d)/2, the semiperimeter of the quadrilateral.
To find the circumradius of a cyclic quadrilateral, first consider the triangle formed by the first three vertices of the quadrilateral. Circumscribe a circle around this triangle. The radius of this circle will be the same as the radius of the circle circumscribed around the cyclic quadrilateral. 
Now reveal the fourth vertex, D, on the circle, and draw
chords from D to each of the other points. Now ABCD is the cyclic
quadrilateral, and AC and BD are its diagonals. The lengths of the sides
are a, b, c, and d, and the lengths of the diagonals are p and q.
Now that I've labeled the figure this way, the circumradius of the quadrilateral is the circumradius of the triangle ABC, R = abp/(4K), where K is the area of triangle ABC. (proof) R = cdp/(4L), where L is the area of triangle CDA. Let A be the area of the quadrilateral. A = K+L R = (abp+cdp)/(4A) = p(ab+cd)/(4A) Similarly, R = q(bc + ad)/(4A), so R^{2} = pq(ab+cd)(bc + ad)/(16A^{2}) Ptolemy's Theorem says ac+bd = pq, so R^{2} = (ac+bd)(ab+cd)(bc + ad)/(16A^{2})
Brahmagupta's Theorem says the area, A, of the quadrilateral is
sqrt((sa)(sb)(sc)(sd)), so 
Circumscribed Triangle  proves the formula for the circumradius of a triangle.
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