Call the length BC = x, which is the radius of circle C2. Then BE is also x, and the base of isosceles triangle ABE, where AB and AE are each 1. Dividing the isosceles triangle into two congruent but opposite right triangles, we see that angle ABE = arccos(x/2). Angle CBE is the complement of this angle, as it's within right angle CBA, so angle CBE = arcsin(x/2).
Triangle CBE is also isosceles, with the apex at B, and we can also divide this isosceles triangle into two right triangles. The angle of such a right triangle at B is therefore arcsin(x/2)/2, and the angle at C is its complement: angle BCE = pi/2 - arcsin(x/2)/2.
Angle BGC is the complement of this, and therefore equal to arcsin(x/2)/2.
As x (which is the radius of circle C2) approaches zero, sin(x/2) approaches x/2 as does tan(x/2), and half of these approach x/4, as do the corresponding angles, in radian measure.
The distance BG is x / tan(BGC), so BG approaches x/(x/4) = 4.
Posted by Charlie
on 2006-01-13 11:05:37