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Explanation 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. 