TITLE: Finding location of a point on 2D plane, given the distances to three other know points
QUESTION [11 upvotes]: I need to find location of the point $s_0$; the locations of other three points ($s_1$, $s_2$, $s_3$) are known. $d_i$ are known distances.
Given: $x_1$, $x_2$, $x_3$, $y_1$, $y_2$, $y_3$, $d_1$, $d_2$, $d_3$
To be found: $x_0$, $y_0$
This will be a physical system. $s_i$ will be fixed antennas and the distances $d_i$ will be read via proximity sensors. The aim will be to make $s_0$ find its own location. So, $s_0$ will always exist; there will always be a unique solution point $s_0:(x_0, y_0)$.
How do I solve this problem?

REPLY [2 votes]: An Alternative to Circles
Since you should get the same answer as with circles, you wind up the same.
The way I understand the original question:
GIVEN:

3 arbitrary points (x,y) that are not colinear. I emphasize that
the X and Y coordinates for all three points are known.
3 distances from the 3 arbitrary points
The fact that the three distances (line segments) share one common point. But we are not given the coordinates of that point. 
We also know the other end of each line segment is one of the 3 arbitrary, non-colinear points. And we even know which segment (distance) goes with each point.

SOLVE:
Calculate the X and Y coordinates of the one common point.
HOW:
First, for the sake of thinking, change the problem. Instead of finding the unknown X,Y of the common point, assume that unknown to be THE KNOWN. And the three distances are 3 UNKNOWNS for which you will solve. This invented problem is easy for you because you know how to calculate the distance between any two cartesian points (X0, Y0) and (X1, Y1) and the Pythagorean theorum namely:

Distance d between two points (X0, Y0) and (X1, Y1):
  d = square_root ((X1 - X0) ^ 2) + ((Y1 - Y0) ^ 2))

Now, switch back to the real problem because you have your relevant formulae. Let the distances be the knowns and the common point (X1, Y1) be the unknown and represented as (X,Y). Let the *k***nown point be represented as **(Xk, Yk). Solve for X and Y. 
After some algebraic manipulations, I think you can get this for each distance:

y = Yk +- square_root (d^2 - X^2 + 2 * X * X + X^2)

where 

+- means plus or minus
^ means the power symbol, such as X^2 meaning X squared



Since you have three different distances, and known points, you have this equation for each. Hence, you have three equations to solve for the two unknowns. "But I thought the rule is you need 2 equations to solve for 2 unknowns." Here you need 3 equations to solve for 2 unknowns because you have a square root. 
In the original problem, intuitive was you need three known reference points and three distances to determine a unique point. Our three equations work like this:
Use any one, you get a curve.
Use any two, you get two points. Two curves can intersect in two points. It's straight lines that can only intersect in one point. We can get 2 answers because we are taking a square root.
You need all three equations to solve for these two unknowns X and Y.
I have not provided the complete answer. Essentially the same method which (unless something changed) you can find someone else previously posted in this thread. I have explained it more.