Math Forum / Mathematics / Undergraduate Math / April 2008

In article
<3640366.1208144047351.JavaMail.jakarta@nitrogen.mathforum.org>,

I suggest an algebraic approach, which may be most easily handled using
complex numbers to represent the points.

There is no loss in generality if we assume that the circle is centered
at 0 and has radius 1.  Then the two points on the diameter of the
circle may be called t and -t.

Let w be the third point.

If z is a variable point, consider its distances from t, -t, and w; z is
the center of the "variable circle" if these distances are equal.  
Express this condition in terms of z, w, t, and their conjugates (I
won't try to type that). Out of these equations, distill a n equation
that is "linear* in z.

I meant to respond to this ages ago, apologies for that. In the  
meantime I have seen one resolution using complex numbers.

An alternative involves a couple of invocations of the Pythagoras  
Therorem - well, 3 I think.

Since you conjecture that the locus is the line perpendicular to the  
join of the point and centre, from one point on the locus drop a  
perpendicular to that line. The locus point can now be seen as the  
common vertex of 3 right angled triangles all having a radius of the  
variable circle for a side. After some invocations of Pythagoras it  
becomes obvious that the foot of the constructed perpendicular is  
invariant, and hence the result.

The situation could be generalised - replace the fixed point by a  
second circle and insist that the variable circle cuts both circles at  
the ends of diameters - and is capable of a similar resolution.

Eves (Harold Eves, College Geometry), calls this locus the anti-
radical axis. I don't know why he uses this term, but the ratio in  
which the line cuts the join of the centres of the circles (the  
invariant foot of the perpendicular, above) is the reciprocal of the  
radical axis ratio.

cheers,

Dave
Dave@zizek.demon.co.uk
http://www.zizek.demon.co.uk
http://www.northendenriversidepark.org.uk