Locus_Concentric

Locus Concentric

Let a and b be two intersecting circles.
Let A be a point on a and B, C be the intersection points of the two circles. Draw AB and AC and produce them until they cut the circle b in two other points D and E.
Consider the circumcenter X of the triangle ADE, as the point A varies on the circle a. The geometric locus of X's is a circle, with radius equal to that of a and center the one of b (i.e. the translate of a by |OP| the distance of centers).

Switch to the pick-move tool (CTRL+2). Pick point A and move it along the circle a.

But let us examine now the fundamental facts underlying the above construction.

1) The angle at C, viewing the segment AD, is constant, say fi, independent of the location of A on a.
In fact, this angle has measure (pi-ang(CAB)-ang(CDB)) and the last two angles have constant measure, since the circles and the chord BC are fixed.

2) A consequence of 1) is that the angle DCE is also fixed, hence the chord DE has constant length, say x, independent of the location of A.

3) The quadrangle BCED is inscribed in b, hence the triangles ABC and AED are similar. Let y=|BC|. The similarity ratio is equal to y/x.

4) To examine this ratio, take A so that AD is parallel to the line OP joining the centers. One sees, by extending CO and CP, that AD has twice the length of OP, and AC, CD are diameters of a and b correspondingly. The angle ang(AED) is a right one, hence by the similarity in 3) y/x = |AC|/|AD| = |OC|/|OP|.

5) A consequence of this is that all triangles AED, for A moving on a, have circumcircles of constant radius, equal to |OP|.
In fact, since the triangles ABC and AED are similar, their circumradii are also proportional with the same ratio = |OC|/|OP|. Since |OC| is the radius of a, |OP| must be the radius of the circumcircle of AED.

6) The equality of the angles ang(ABC) = ang(AED) implies that the triangles AOC and AQD are similar isoscelii, whose all base-angles are equal, say to w.

7) Now consider the center Q of the circumcircle of AED and the point S of intersection of BC with the radius QA. The angle ang(ASC) is a right one.
In fact, ang(ASC) = pi-ang(SAC)-ang(ACS). But ang(SAC) = ang(BAC)-w, whereas ang(ACS) = ang(OCS)+w. Thus ang(ASC) = pi - ang(BAC) -ang(OCS), wich is a right one.

8) A consequence of 7) is that (the radius of the circumcircle of AED) QA is parallel to OP. Beeing also equal to this in length, the quadrangle AOPQ is a parallelogram, thus |AO| = |QP|. This proves the proposition at the top of the document and justifies the figure drawn there.

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