ThreeCirclesProblem1

Three circles problem (related conic)

(Continuation from ThreeCirclesProblem.html )
Let a = (O, OB), b = (C, CB) and c = (D, DE) be three circles. (a) and (b) are tangent at B. All three have fixed radius and B moves on (a), remaining tangent to it all the time. For each position of B the bisectors of angle AOB are drawn and their intersections F, G with the radical axis of the circles (b) and (c) are defined. F and G move on two lines orthogonal to the common diameter AD of circles (a) and (c). The circumcircle (d) of the triangle FOG passes through three other interesting points. These are: H the intersection of JN with AO, JN being the symmetric of radius OB of (a) with respect to the line [FG]. (d) passes also through I and J, symmetric of H and O, with respect to [FG], respectively.

Assuming that F, G project always on the fixed points K, L of AO, respectively, we consider the segment BN, orthogonal to FG and parallel to CD, which joins the centers of circles (b) and (c). Let M be the intesection point of BN and AO. From the equality OM/OD = OB/OC we see that M is fixed on AO. Because FG is a diameter of (d), J, the symmetric of O, with respect to that diameter is on (d) too. An easy calculation shows that angle(OHJ) = angle(OFJ), hence H is also on the circle (d). Lastly, I is on (d) for the same reason (J) is. The projection S of the center R of (d) is the middle of KL and H is symmetric to O with respect to S. This implies that H is fixed on AO.
Since JN and OB have same length and HJ/HN = HM/HO, the length of HJ (and of HN) remains constant. This has a nice consequence.
Point P is the intersection of the two segments OB and JN, symmetric with respect to FG. Since the total length PH + PO = HJ is constant and points H and O are fixed, point P is on the ellipse (e) with foci at H and O and great axis half the length of HJ. Besides line FG is tangent to the ellipse at its point P. Since lines FK and GL are the particular tangents to that ellipse, which are orthogonal to its great axis, the distance HJ is equal to KL and equal also to the radius of the [major] circle of the ellipse.
Notice that the ellipse (e) becomes a parabola/hyperbola, depending on the positions/dimensions of the three involved circles. Besides, there are several other remarkable curves related to this figure. Look at ThreeCirclesProblem2.html , if you are interested in.

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