Consider a circle c(H,r) orthogonal to another circle b(A,R). Draw a diameter d = CD of (b) and the line (a) orthogonal to CD through H. Let F be the intersection point of (a) with CD. Show that the power of F w.r. to (c) is FK*FL = FE^2. Thus it is independent of the location of H on (a), provided (c) is orthogonal to (b).
The proof is a simple calculation: FK*FL = FG^2 = FH^2 - GH^2 = FH^2 - HI^2 = FH^2-HE*HJ = (1/4)(HJ+HE)^2 - HE*HJ = FE^2 = > FE = FG.
The figure has also another, much more structural, interpretation. Circle (b) can be considered to belong to bundle II(b,a) generated by (b) and (a). This is a bundle of intersecting type, which is orthogonal to a second bundle I(c, d) generated by (c) and (d). Then (d) is the common radical axis of all circle-members of bundle I(c,d) and consequently a point on the radical axis (and not only the distinguished F) has the same power w.r. to all circle-members (c) of bundle I. FE is the radius of the minimal circle-member of bundle II(b,a).
The property generalizes to the following. For any D on the line (d), considered common radical axis of all circle-members of bundle I, DE^2 = DM*DN. This is again a consequence of the orthogonality of bundles. Since DM*DN is the power of point D w.r. to (c) and the circle f(D, DE) is orthogonal to I.
See the file Orthodiagonal.html for an interesting application of this exercise.
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Orthogonal Bundles ![[0_0]](OrthogonalBundles_files/OrthogonalBundles_0_0_0.png)
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