HYACINTHOS 4622

• On Wed, 2 Jan 2002, Milorad Stevanovic wrote:
  
  

  > Paul wrote :
  > > [The] inversive image of the nine-point circle in the
  > >radical circle of the excircles [has] radius (r^2+s^2)/(4r).
  > > [The] center[,] X(970)[,] lies on the Brocard axis.
  > > Is there a simple explanation of this fact?
  
  

  [Jean-Pierre(?):]
  
  

  > [...] Thus your circle is a Tucker circle and he is centered on OK.
  > In fact, as the radical center of the excircles is the Spieker point
  > S, the center of your circle is the common point of OK and SN
  > (N=NPcenter), which is, as you've noticed X(970).
  
  

  I suspect there must also be some connection with "the Conway circle"
  (center Io), since the radius of that is root(rr+ss). There's probably
  a vast generalization that describes all the Tucker circles in some
  such way (which maybe Jean-Pierre is quoting).
  
  John Conway