Intersert With Breaking at points of intersection

[Steve]

I'm not sure exactly what you're asking for, but if it's how to render the
two Arcs from a split circle, check out the Circle() method and the Start
and End parameters. Your single circle will essentially become two calls
specifying starts and ends for each arc.

Steve.

"Greg Schmidt" <gschmidt@cablesystemservices.com> wrote:
>Help,
>
>I am trying to break a circle at any point that intersects a line. I have
>the pointsusing the intersect with statement, but don't know how to break
>the circle.
>
>Thanks,
>Greg
>
>

[Greg Schmidt]

Help,

I am trying to break a circle at any point that intersects a line. I have
the pointsusing the intersect with statement, but don't know how to break
the circle.

Thanks,
Greg

[Greg Schmidt]

I just need to break a circle or an arc. I have the points but don't know
the statement to do this.

"Steve" <stevepyn@hotmail.com> wrote in message
news:3d85f6e1$1@10.1.10.29...
>
> I'm not sure exactly what you're asking for, but if it's how to render the
> two Arcs from a split circle, check out the Circle() method and the Start
> and End parameters. Your single circle will essentially become two calls
> specifying starts and ends for each arc.
>
> Steve.
>
> "Greg Schmidt" <gschmidt@cablesystemservices.com> wrote:
> >Help,
> >
> >I am trying to break a circle at any point that intersects a line. I
have
> >the pointsusing the intersect with statement, but don't know how to break
> >the circle.
> >
> >Thanks,
> >Greg
> >
> >
>

[Russell P. Holsclaw]

"Greg Schmidt" <gschmidt@cablesystemservices.com> wrote in message
news:3d85f8f5$1@10.1.10.29...
> I just need to break a circle or an arc. I have the points but
don't know
> the statement to do this.

This is only a wild guess, but it sounds like this question pertains
to using VBA within the context of a graphics editor, probably Visio.
If that's the case, you may want to post your question to the
appropriate newsgroup, namely "microsoft.public.visio.developer.vba".

Hope this helps.

--
Russ Holsclaw