Consider a circle O with a diameter AB, shown here in green. Draw a second circle (red) with diameter AC, such that C is on AB. Draw an isosceles triangle with base CB and third vertex D on circle O. Draw a third circle (X), tangent to the first three figures. Prove that the line from C to the center of circle X is perpendicular to AB.
Let the diameters of circles AC and AB be 2a and 2b respectively. Let M be the midpoint of line segment CB. If we let C be the origin (0,0), then we have A = (2a,0), B = (2b2a,0), and M = (ba,0). Circle AB has the equation
(x+2ab)^2 + y^2 = b^2.
Plugging the xcoordinate of D ( =ba ) into this we get
D = (ba,sqrt(b^2a^2)).
If circle X has radius r, then its center X = (x,y) must be one of the intersections of the circles:
(x+a)^2 + y^2 = (a+r)^2 and (x+2ab)^2 + y^2 = (br)^2.
Solving for r we get
r = (ba)(2a+x)/((a+b).
Since we want the center to be on the yaxis ( x=0 ),
r = 2a(ba)/(a+b).
Solving for y gives
y = 2a*sqrt(2b(ba))/(a+b).
Now we must show that circle X is tangent to line CD.
Let T be a point on circle X such that CT is tangent to circle X and T is on the same side of line CX as point D. Clearly circle X is tangent to line CD if triangles CXT and DCM are similar.
Both are right triangles with
CM = ba and DM = sqrt(b^2a^2)
XT = 2a(ba)/(a+b) and XC = 2a*sqrt(2b(ba))/(a+b).
With these four values, I will let you verify that the triangles are similar.

Posted by Bractals
on 20040910 16:18:51 