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Look into the ATTOUT command.

I think that is what happened in the first place. thanks a lot Steven. Cheers

You may want to try changing the compatibility mode for your autocad to the appropriate release of windows for the acad program.

Your structure of the list is the culprit and not the boundary command. I did not get through you codes but I am correcting the statement that you alluded to. (COMMAND "-BOUNDARY" (list (+ (CAR startPoint) (/ width 2.0)) (- (CADR rectEnd) 100.5)) "")

It sounds like you are working in a locked viewport and panning (moving?). If so, AutoCAD needs to switch to paper space while panning and then switch back to model space to execute the command.

I bet BIGAL will see this and get back to you pretty quick, he has posted before about similar issues and solutions.

in some cases the viewport layer is off so it isn't clear there is one. Might be part of it. As long as you double click into the viewport it will switch into model space. You may need to lock the viewport if you don't intend to change the scaling. -ChriS

Found it Circle 3p "Tan" pick 1st line pick pt "Tan" pick 2nd line So can use this to get radius then fillet.

You will probably need to provide more information. But, a first guess is you are in paperspace and trying to select something in model space

If I understand your goal you want to create a fillet between 2 straight lines such that the fillet arc passes through a specified point "p" that is not on either of the two lines. The radius "r" of the arc is unknown. If so, you can draw the angle bisector to the two lines. There are actually two angle bisectors but we can assume the one to use is implied by the bounds of the lines. The center of the fillet is on this line but where? We know that arc center is a distance r from either line (measured perpendicular to the line) and from the point p. You can guess a point p1 on the angle bisector and then determine r and the distance s (the distance from p1 to p). If s is less than r determine a new point p2 further out from the intersection of the two lines. and recalculate r and s. Repeat until r and s are equal within a given tolerance. You can modify the guessing process to speed convergence to a solution.

Is it assumed they specified point is on a curve that can even be tangent to the two lines? For instance, the point shown in the attached image cannot exists on a curve that is tangent to the two lines.