Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

I would like to draw a quadrilateral inscribed within a circle. How can I construct this figure, taking into account arbitrary (specified) side lengths, while still ensuring that the vertices of the quadrilateral lie on the circle?

share|improve this question
It is to be quad in a circle with a user-specified length of the side of a quad, is to change the circle to fit a quad in a circle. –  user5955 Feb 18 '13 at 13:12
Just that, given the length of the side of the square will fit within a circle, which is automatically size to fit the square. –  user5955 Feb 18 '13 at 13:21
I have edited your question according to what I think it means. Can you confirm whether or not this is correct? –  Oleksandr R. Feb 18 '13 at 13:24
Perhaps mathworld.wolfram.com/CyclicQuadrilateral.html ? –  cormullion Feb 18 '13 at 13:37
@user5955 You image showed a rectangular inscribed. Therefore, I assumed you don't need an arbitrary quadrilateral. –  halirutan Feb 18 '13 at 13:52

1 Answer 1

up vote 5 down vote accepted

I hope I understood your question correctly. When you place your figure at {0,0}, meaning the center of the circle and the center of the rectangle is there, you don't need to calculate very much. Indeed, everything is then fixed by exactly one point p defining a corner of the rectangle and the radius of the circle.

A dynamic version of your graphics can be written down in only a few lines of code

 Graphics[{FaceForm[None], EdgeForm[Thick], Rectangle[-p, p],
   Thick, Red, Circle[{0, 0}, Norm[p]]}, PlotRange -> {{-2, 2}, {-2, 2}}],
 {{p, {1, 1}}, Locator}

Mathematica graphics

share|improve this answer
Yes thanks for your help. Best regards. –  user5955 Feb 18 '13 at 13:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.