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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?

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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 ? – 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
up vote 6 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

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Yes thanks for your help. Best regards. – user5955 Feb 18 '13 at 13:54

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