534 
Proceedings of the Royal Society 
where & and 6 are given by 
(b - d) 2 = a 2 -\- c 2 - 2 ac cos 0 , 
(b + d) 2 = a 2 + c 2 - 2ac cos O'. 
The expressions for the other chords differ only by the inter- 
change of b and c. Elimination gives at once 
P 
» o_-.fl q 2 + c 2 -(6-rf) 2 a 2 + c 2 -(6 + cQ 
2ac 
2 ac 
^ | i ( a 2 + c 2 -(b-d) 2 y |*| 1 _( a* + <*-(b + d)* y 
= J_ { 4 (a? + d 2 )(b 2 + <?) - 2(6 2 - c 2 ) 2 - 2(a 2 - <Pf 
4 a 2 { 
T 2(4a 2 c 2 - (a 2 + c 2 - (b - rf) 2 ) 2 )i(4a 2 c 2 - (a 2 + c 2 - (b + d) 2 ) 2 ) i l 
= ^ ^A 2 + A' 2 t 2AA'^ 
where A and A' are the areas of the “ inscribahle” quadrilaterals, 
crossed and uncrossed, whose sides are a, b, c } d. This, of course, 
proves Talbot’s theorem. 
Hence 
a remarkably simple expression. The two values of p are given at 
once by Talbot’s diagram, and the rectangles under their quarter 
sum, and difference, respectively, with the distance between the 
centres, give the areas of the quadrilaterals above mentioned. Or, 
better, the triangles whose angular points are the middles of the 
arcs respectively, and the centres, have areas equal to half the sum 
and half the difference of the quadrilaterals. 
The symmetry of these expressions shows that in Talbot’s theorem 
any two of the four quantities employed may be interchanged — the 
lengths of the corresponding pairs of equal chords being always 
inversely as the quantity chosen for the distance between the centres 
of the two series of circles. 
