# add a point, close to one but distant to the others

I am given a finite set of geometric points in three dimensions. I want to add another point $A$, so that it's close to a certain point $P$ (that is $d(P, A)\lt k$, where k is an arbitrary constant), but distant from the others ($d(P, X)\gt k$, where $X$ is any other point).

I am clueless. Numerical solutions or fallible algorithms are welcome.

One simplification I thought is that one could look for a certain surface passing through $P$, so that the other points are all on one side of the subdivided space, then place $A$ to the other side. This can cut off some solutions to the first problem, but since in my actual situation the point positions have some uncertainty, a stricter solution leaves the margin of error that I need. Unfortunately, I didn't make up a good method to calculate this surface (rather a set of) which solve the problem.

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Is your set of points finite? –  you Mar 22 '12 at 23:36
yes. see edit . –  Lorenzo Pistone Mar 22 '12 at 23:37
Ah ok, good :) Well, for starters you can find a point arbitrarily close to $P=(x_1,x_2,x_2)$ by taking $A=(x_1+\sqrt{\frac{\epsilon}{3}},x_2+\sqrt{\frac{\epsilon}{3}},x_3+\sqrt{\frac{‌​\epsilon}{3}})$ for any $\epsilon>0$, so that $d(A,P)=\epsilon$. Then it's just a question of exactly how far away you want $A$ to be from the other points. –  you Mar 22 '12 at 23:41
That equation makes the new point $A$ to be fixed on a certain line, as the parameter $\epsilon$ is only one. The new point is not subject to such constraint. –  Lorenzo Pistone Mar 22 '12 at 23:43
In that case, you could take any vector $v\in\mathbb{R}^3$ of length $\epsilon$ and let $A=P+v$ –  you Mar 22 '12 at 23:50