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March 2nd, 2018, 09:16 AM  #11 
Senior Member Joined: Nov 2011 Posts: 197 Thanks: 2 
Thanks!!! I think that the question will be solve by computer. Right?! 
March 2nd, 2018, 10:06 AM  #12  
Senior Member Joined: Sep 2015 From: USA Posts: 1,980 Thanks: 1027  Quote:
It would be a better problem if $x,y \in \mathbb{R}$ as this would allow an exact answer. Since you imply a uniform distribution of points we simply have to find the radius such that the areas of the two regions are equal. i.e. $k^2  \pi r^2 = \pi r^2$ $k^2 = 2 \pi r^2$ $r = \sqrt{\dfrac{k^2}{2\pi}} = \dfrac{k}{\sqrt{2\pi}}$  

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