--- 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.