Find the x,y location of one point (unknown location) in space with respect to four other known points There have been many other questions like this one but they involve 3 points instead of four. I have code that will find a transmitter (red-$X$) in between 3 different points and it works great. I was hoping to add another receiver or in other words a 4th outside point. Is there any ideas of how I can modify this function to find the red-$X$ ($x,y$ form) in the plot with respect to four blue points instead of just three? (see plot or plot link below) Even just the Mathematics behind finding a point (having unknown coordinates) with respect to four other fixed points (known coordinates) would be great! Thank you very much.
Solving nonlinear equations with fsolve
% pylab inline
import pylab
from scipy.optimize import fsolve

def equations(p):
    # Define these outside of the function before calling this function.
    global gamma01,x0,y0,gamma12,x1,y1,x2,y2,gamma10
    x,y = p
    # The returned equations are from Power ~ 1/r**2, so
    # the power ratio gammajk = Pj/Pk = rk**2/rj**2.
    return ( gamma01*(x1-x)**2+gamma01*(y1-y)**2-(x0-x)**2-(y0-y)**2,
             gamma12*(x2-x)**2+gamma12*(y2-y)**2-(x1-x)**2-(y1-y)**2 )

gamma01 = 1.0  # Received power antenna 1 over received power antenna 0
gamma12 = 1.0  # Received power antenna 2 over received power antenna 1

x0,y0 = 0.0, 1000.0  # Position receive antenna 0
x1,y1 = 1000.0, 0.0  # Position receive antenna 1
x2,y2 = 0.0, -1000.0 # Position receive antenna 2

# Numerically solve our nonlinear system of equations
# (1.0,1.0) is the initial guessed position
x, y =  fsolve(equations, (1.0, 1.0))
print('answer x y (m)',x,y)

pylab.figure()
pylab.plot([x0,x1,x2],[y0,y1,y2],'bo',markersize=8.0,label='Receive Antenna')
pylab.plot([x],[y],'rx',markersize=8.0,label='Transmitter')
pylab.axis('equal')
pylab.xlabel('x (m)')
pylab.ylabel('y (m)')
pylab.title('All Power Ratios = 1.0')
pylab.legend()
pylab.grid()
pylab.show()

Plot of red-X with respect to Blue dots
 A: Finding the unknown location having as data known distances to three known points is a matter of intersecting three circumferences. There may be none, one or two solutions.
If you have four points then calculate each set of three points, and then calculate the average of the solutions, if they exist. It's not very "mathematical", but gives you a good approximation of the unknown point.
A: Let us make the problem general with $n$ receivers of known coordinates $(x_i,y_i)$ and a a transmitter of unknown coordinates $(X,Y)$ and let $d_i$ to be the distances.
So the real problem is to minimize with respect to $X$ and $Y$ the function
$$F=\frac 12\sum_{i=1}^n \left(\sqrt{(X-x_i)^2+(Y-y_i)^2}-d_i\right)^2$$ and this will require good initial estimates.
To get these estimates, consider in a preliminary step the $n$ equations
$$f_i=(X-x_i)^2+(Y-y_i)^2-d^2_i=0$$ and build the $\color{red}{\frac {n(n-1)}2}$ equations
$$g_{ij}=f_i-f_j=2(x_j-x_i)X+2(y_j-y_i)Y+\Big[(x_i^2+y_i^2-d_i^2)-(x_j^2+y_j^2-d_j^2)\Big]=0$$
that you can easily solve using multilinear regression with no intercept (or matrix calculations). You could also notice that the minimum of
$$\sum _{k=1}^n ( a_k X+ b_k Y+c_k)^2$$ is obtained solving
$$X \sum _{k=1}^n a_k^2 + Y \sum _{k=1}^n a_kb_k+\sum _{k=1}^n a_kc_k=0$$
$$X \sum _{k=1}^n a_kb_k + Y \sum _{k=1}^n b_k^2+\sum _{k=1}^n b_kc_k=0$$
This will give you the estimates of $X$ and $Y$.
If you want to polish the solution, going back to $F$, you need to solve the equations
$$\frac{\partial F}{\partial X}=\sum_{i=1}^n \frac{(X-x_i) \left(\sqrt{(X-x_i)^2+(Y-y_i)^2}-d_i\right)}{\sqrt{(X-x_i)^2+(Y-y_i)^2}}=0$$
$$\frac{\partial F}{\partial Y}=\sum_{i=1}^n \frac{(Y-y_i) \left(\sqrt{(X-x_i)^2+(Y-y_i)^2}-d_i\right)}{\sqrt{(X-x_i)^2+(Y-y_i)^2}}=0$$ which will be solved using Newton-Raphson method. If you are lazy as I am, do not waste time establishing the required derivatives $\frac{\partial^2 F}{\partial X^2}$,$\frac{\partial^2 F}{\partial X\partial Y}$, $\frac{\partial^2 F}{\partial Y^2}$ and just use central differences to get them numerically. Since the preliminary step will give you good estimates, this would converge very fast.
