Posted by dxli on Mar 13, 2015; 8:07pm
URL: https://forum.librecad.org/Isometric-Circles-tp5711165p5711188.html

Interesting enough.

To find the inscribed ellipse of a symmetric trapezoid can be converted to find roots of a quartic equation.

The general algorithm by projection of inscribed circle of square doesn't apply here.

There's no solution for asymmetric trapezoid.

stranger573 wrote

Mike Hayes wrote

...
So I'm guessing the code behind the tool is already doing something extra to create a unique solution. Perhaps it is making the ellipse touch the midpoints of the four selected lines?
...

The area of the inscribed ellipse (and other objects) should be the maximum. For this there is only one solution.
I believe that this is a solution for isometric circles too.

Midpoint is not good idea. Example: