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I have a 2 points i know the distance between them with a 3rd points. how to get the position of the 3rd point given other distances with the other two points ?

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Given only this information, you generally can't know the specific point. That third point will be at an intersection of two circles--one centered around each of the other two points. In particular, let's call your first two points $a,b$ and the point you're looking for $c.$ Let $r_{ab}$ be the distance from point $a$ to point $b$, and let $r_{ac},r_{bc}$ be the desired respective distances from $a,b$ to $c$. What we're dealing with, then, is circles around $a$ and $b$ of respective radii $r_{ac}$ and $r_{bc}$.

Now, if two of $r_{ab},r_{ac},r_{bc}$ add up to the third, then there is exactly one point of intersection of the two circles--and that is our point $c$. If two of $r_{ab},r_{ac},r_{bc}$ add up to some number that is less than the third, then the circles don't intersect at all, and there is no such point $c$. Otherwise, there will be two points of intersection of the circles, and $c$ will be one of those, but (as I said) there's no way to know which, given only this information.

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Well, assuming that we're in a plane (as I do implicitly in my answer), let $a=(h_a,k_a)$ and $b=(h_b,k_b).$ The point $c$ that we're looking for will be a solution to the following system of equations (if any exist): $$\begin{cases}(x-h_a)^2+(y-k_a)^2=r_{ac}^2 & \\(x-h_b)^2+(y-k_b)^2=r_{bc}^2 & \end{cases}$$ As mentioned in my answer, this system of equations will have either $0,1$ or $2$ solutions. As a side note (if you're wondering how $r_{ab}$ fits into this), we have $$(h_a-h_b)^2+(k_a-k_b)^2=r_{ab}^2.$$ –  Cameron Buie Oct 30 '13 at 3:18
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Honestly, the solution of the general equation is really ugly. Expanding the squared binomials and subtracting one equation from the other, we see that $$2(h_a-h_b)x+2(k_a-k_b)y+h_b^2-h_a^2+k_b^2-k_a^2=r_{bc}^2-r_{ac}^2.$$ Of course if we have $h_a=h_b$ or $k_a=k_b,$ this becomes fairly simple to solve for whichever of $x,y$ isn't eliminated, at which point we simply substitute back into one of the original equations and solve for the other. Even if not, we can solve readily. Regardless, though, the formulae for the general solutions will be nasty. –  Cameron Buie Oct 30 '13 at 3:33
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After substitution, we will end up with a quadratic in $x$ or $y,$ which is why we can only have $0,$ $1,$ or $2$ solutions. –  Cameron Buie Oct 30 '13 at 3:35
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Well, if you're assuming that, then the task is made easier. The system becomes $$\begin{cases}x^2+y^2=r_{ac}^2 & \\(x-r_{ab})^2+y^2=r_{bc}^2, & \end{cases}$$ which is much easier to solve. Can you take it from that point? (The strategy will be much the same, regardless.) –  Cameron Buie Oct 30 '13 at 4:08
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As I said, there will not be a nice answer in general. With your very big assumptions, you get a somewhat nice answer, but usually, it's nasty-looking. –  Cameron Buie Oct 30 '13 at 4:23
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