Hi Kaye,
I redrew the diagram you sent based on out email conversation, and labeled some points.
I drew a coordinate axis with the origin at O, the midpoint of the line of length B. V is the centre of the circle and lies of the yaxis. Since P and Q are on the circle the centre must lie on the right bisector of the line segment PQ. Hence I am going to find the equation of this right bisector and see where it crosses the yaxis. This will give me the centre and radius of the circle from which you should be able to determine the other dimensions you need.
P has coordinates (0,G) and Q has coordinates ( ^{B}/_{2} + C + A, D) and hence S the midpoint of PQ has coordinates
( ^{(B + 2C + 2A)}/_{4} , ^{(D + G)}/_{2})
The line through P and Q has slope
^{(2D  2G)}/_{(B + 2C + 2A)}
and thus the right bisector of the line segment PQ has slope
^{(B + 2C + 2A)}/_{(2G  2D)}
Thus the right bisector of the line segment PQ has equation
y  ^{(D + G)}/_{2} = ^{(B + 2C + 2A)}/_{(2G  2D)} (x  ^{(B + 2C + 2A)}/_{4} )
This line intersects the yaxis at V which has xcoordinate 0 so the ycoordinate of V is
y = ^{(B + 2C + 2A)}/_{(2G  2D)} (0  ^{(B + 2C + 2A)}/_{4} ) + ^{(D + G)}/_{2}
=  ^{(B + 2C + 2A)2}/_{(8G  8D)} + ^{(D + G)}/_{2}
With the example dimensions you sent I got V to have coordinates (0, 55.12) and hence the radius of the circle is 23.5 + 55.1 = 78.6".
Harley
