Fixed Point of Circles Orthogonal to the Given One: What is this about?
A Mathematical Droodle


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Explanation

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Copyright © 1996-2018 Alexander Bogomolny

R. Honsberger credits Hiroshi Haruki with the following unexpected result:

  Given a circle C with center O and a line m, not intersecting C. There exists a point Q such that, for every P on m, PQ equals the length of the tangent from P to C.

In other words, a circle centered at P with the radius equal to the length of the tangent from P to C passes through a fixed point Q. Q is a point of concurrency of all circles orthogonal to C and center on m!


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The proof requires a few applications of the Pythagorean proposition.

Let M be the foot of the perpendicular from O onto M, MN and PT tangent to C, r = ON = OT. Assume also that Q is chosen on MO such that MQ = MN = z. The Pythagorean propositions will be used in the right triangles OPT, OMP, QMP, and OMN:

(OPT)r2 + PT2 = OP2
(OMP)OM2 + MP2 = OP2
(QMP)z2 + MP2 = PQ2
(OMN)r2 + z2 = MO2

From the first two we get

(1)r2 + PT2 = OM2 + MP2.

The second pair yields

(2)OM2 + MP2 = r2 + PQ2.

Now a comparison of (1) and (2) yields PT2 = PQ2. Hence, PT = PQ.

(Note that, in general, x2 = y2, does not imply x = y. Following such uncritical implication may lead to incorrect results. In our case, though, we know up front that both quantities PT and PQ are positive; they both denote the lengths of two line segments. The implication then becomes unequivocally correct: if the squares of two positive numbers are equal the numbers themselves are also equal.)

Remark

It is surprising how the same fact acquires an aura of familiarity if looked at from a different angle. In a discussion that involves orthogonality of circles, one thing that most certainly comes to mind is the coaxal circles theorem: circles in an Apollonian family are all orthogonal to circles through two fixed points (and vice versa.) The circles in the latter family have their centers on a fixed straight line - the radical axis of any two circles from the Apollonian family.

Thus it is obvious that, along with Q, there is a second point common to all circles with the center on m orthogonal to C. This point is the reflection of Q in m.

In the spirit of the above (see also the discussion on the Apollonian circle), Nathan Bowler suggested to consider circles orthogonal to two non-intersecting circles C and m, which makes the situation more transparent:

Let Q be the intersection of any pair of circles orthogonal to both C and m. Invert with respect to Q. Then these two circles invert to a pair of intersecting lines. Call the point of intersection R. The inverse of C is a circle perpendicular to both of the lines, and so with centre on both lines. That is, it is a circle centred at R. Similarly the inverse of m is a circle centred at R. Then any circle perpendicular to both m and C must invert to a circle or line perpendicular to each of the two concentric circles, that is, to a line through their common centre at R. So any circle orthogonal to both C and m must pass through both Q and R', the inverse of R.

This proof shows that there are 2 such fixed points. Note that I have nowhere used the fact that m is a straight line, so that the proof holds for a general pair of nonintersecting circles.

References

  1. R. Honsberger, The Butterfly Problem and Other Delicacies from the Noble Art of Euclidean Geometry II, TYCMJ, 14 (1983), pp. 154-158.

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