[pstricks] {pst-eucl?} Drawing a circle with tangent

[Hubert Lam]

Hi again all

No need to worry -- found the solution in the other example. Just draw
another circle! :0


\pstGeonode(0,0){O}(3,0){M}
\pstCircleOA[Radius=\pstDistVal{2}]{O}{}
\pstMiddleAB[PointSymbol=none, PointName=none]{O}{M}{O'}
\pstInterCC[RadiusA=\pstDistVal{2}, DiameterB=\pstDistAB{O}{M},
                                CodeFigB=true, CodeFigColor=magenta,
                                PosAngleB=45]
                      {O}{}{O'}{}{A}{B}
\psset{linecolor=red, linewidth=1.3\pslinewidth, nodesep=-2}
\pstLineAB{M}{A}\pstLineAB{M}{B}

Thanks

Hubert


On Mon, 02 Aug 2010 08:35 +0200, "Herbert Voss"
<Herbert.Voss at fu-berlin.de> wrote:
> Am 02.08.2010 04:55, schrieb Hubert Lam:
> > In the documentation of pst-eucl the author shows how to draw a circle
> > with a tangent by specifying three points: the end points of the tangent
> > and the centre of the circle, then the point of contact is found by
> > \pstProjection (i.e. radius is perpendicular to point of contact) and
> > then the circle is drawn using \pstCircleOA.
> 
> Hubert,
> to which page of the documentation do you refer?
> 
> > Is there a way of using the macros in pst-eucl (and/or pstricks-add) to
> > specify the radius of the circle, and then drawing the tangent from
> > there without having to differentiate the semicircle function, find the
> > coordinates of the point of contact and then substitute into y = mx + b?
> 
> Herbert
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