geometry app question...

[webmasterpdx]

I'm trying to draw a circle, tangent to 3 lines using incircle().

I have 3 lines defined as:
GA: line(y=(2*x+5))
GB: line(y=(-2*x-4))
GC: line(y=(-7*x+5))
and then I tried incircle as in:
GD: incircle(GA,GB,GC)

It draws the lines fine, but there is no circle on the plot screen....
What am I doing wrong?
Thanks
-Donald

[Arno K]

On-calc help says incircle uses 3 Points rather than 3 lines.
Arno

[webmasterpdx]

You are right....but that's kinda screwed up, as it's supposed to draw a circle that is a tangent to 3 sides of a triangle. When I specify 3 points, I don't see the edges of the triangle....just the vertices and a circle in the middle somewhere....i.e. You can't see the tangents.

[Didier Lachieze]

Starting from your 3 lines you can then define the triangle vertices and draw the circle :
GA:= line(y = (2*x+5));
GB:= line(y = (-2*x-4));
GC:= line(y = (-7*x+5));
GD:= single_inter(GA,GB);
GE:= single_inter(GB,GC);
GG:= single_inter(GA,GC);
GH:= incircle(GD,GE,GG);

If you have just 3 points GA,GB,GC you can draw the triangle with triangle(GA,GB,GC)

[webmasterpdx]

Cool! Thank you!
-D

[Brad Barton]

(10-15-2017 07:06 AM)Didier Lachieze Wrote:  Starting from your 3 lines you can then define the triangle vertices and draw the circle :
GA:= line(y = (2*x+5));
GB:= line(y = (-2*x-4));
GC:= line(y = (-7*x+5));
GD:= single_inter(GA,GB);
GE:= single_inter(GB,GC);
GG:= single_inter(GA,GC);
GH:= incircle(GD,GE,GG);

If you have just 3 points GA,GB,GC you can draw the triangle with triangle(GA,GB,GC)

I tried this out today, and it looks pretty good in the plot screen. However when I try to generate the coordinates of the intersection of the circle and any one of the tangent lines, I get an imaginary result. Does this mean that the circle and the lines don't truly intersect?

[Arno K]

If I remember correctly, points in geometry are represented by complex numbers.
Arno

[Brad Barton]

(10-16-2017 08:25 PM)Arno K Wrote:  If I remember correctly, points in geometry are represented by complex numbers.

Thanks, that explains it.