## [EMHL] Re: Line construction problem

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• ... (P,s). ... Dear Antreas, As circle M* rolls around N, the base of the symmetry line moves along a pedal circle. When symmetry line passes through point P
Message 1 of 6 , Jan 23, 2004
--- In Hyacinthos@yahoogroups.com, "Antreas P. Hatzipolakis"
<xpolakis@o...> wrote:
> [APH]:
> >> >Let P be a fixed point, and (M,R), (N,r) two given circles.
> >> >
> >> >Draw a line l through P such that the reflection (M*,R) of
> >> >the circle (M,R) in the line l be tangent to the circle (N,r).
> >>
> >>
> >> Now, instead of the zero circle (P,0) we are given the circle
(P,s).
> >>
> >> Draw a line tangent to (P,s) such that....
>
> [Rafi]:
>
> > Find the homothety centers O' and O" of circle P with each of
> > the two pedal circles corresponding to internal and external
> > tangency of M* and N.
> >
> > Construct circles C1 and C2 on diameters MO' and NO".
> >
> > The intersections of the C1 with two pedal circles and
> > the intersections of C2 with two pedal circles are the
> > candidates to form together with points O' and O"
> > the resulting symmetry axis tangent to circle P.
> >
> >
> > Friendly,
> >
>
> Dear Rafi.
>
> Probably only Darij could decipher the above !
>
> The general problem can be solved by the intersection of
> a circle and a cubic.
>
> Antreas
> --

Dear Antreas,

As circle M* rolls around N, the base of the symmetry line moves
along a pedal circle. When symmetry line passes through point P
the right angle made by the symmetry line and the center line MM*
is inscribed in a circle that has segment MP (NP) as diameter.
The cross-points of these circles give us the second point to
construct line in question. There are two pedal circles -
for internal and external tangency of M* and N.

When P is a circle, it is reduced to a point at it's
homothety centers with 2 pedal circles.

Friendly,
Rafi.
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