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I have four points (say $P_1$, $P_2$, $P_3$ and $P_4$) in $2$-dimensional euclidean space. These four points form a convex quadrilateral.

Now lets say that $P_1$ and $P_3$ are opposite, and $P_2$ and $P_4$ are opposite. This means that the line segment $P_1 P_3$ makes a diagonal $d_1$, and the line segment $P_2 P_4$ makes a diagonal $d_2$.

Now further assume $|d_1|$ and $|d_2|$ are not the same (say $|d_1|>|d_2|$). Let’s assume that $d_1$ and $d_2$ intersect at a point $P'$ in such a way rthat the length of the line segment $P_4 P'$ is smaller than the length of $P'P_2$. This means the length of the line segment $P_4P'$ is strictly smaller than $|d_1|/2$ (as $|d_1| > |d_2|$ implies $|d_1|/2>|d_2|/2$, and $|d_2|/2>|P_4 P'|$).

Now a circle with radius $|d_1|/2$ and centered at $\tfrac{1}{2}(P_1 + P_3)$ will have $P_1$ and $P_3$ at its boundary. Can we say anything that a circle will also include $P_4$ in the above setting? I think we can, but I can’t make an argument.

Are Let me say it more precisely can we give two circles in above setting such that one include opposite points $P_1$ and $P_3$ but not any of the other two points. And the other circle includes $P_2$ and $P_4$ but not any of the other two points. Actually I think we cannot give such two circles why?

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Is it enough to look at counterexample? enter image description here

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  • $\begingroup$ awesome counter example, I have made an edit to my question, actually in above counter example now you cannot a circle that contains P3 and P4 but not P1. $\endgroup$ Jan 15, 2015 at 10:59

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