# Introduction

Imagine you are on a two dimensional cartesian plane and want to determine your position on it. You know 3 points on that plane and your distance to each of them. While it is always possible to calculate your position from that, doing that in your head is pretty hard. So you decide to write a program for that.

# The Challenge

Given 3 points and your distance to them, output the cordinates of your position.

• Input and output may be in any convenient format, including using complex instead of real numbers. Please clarify in your answer which format you use.
• You will always get exactly 3 distinct points with their distance to you.
• The coordinates and distances will be floats with arbitrary precision. Your output has to be correct to 3 decimal places. Rounding is up to you. Please clarify in your answer.
• You may assume that the three points are not collinear, so there will be always an unique solution.
• You are not allowed to bruteforce the solution.
• You may not use any builtins that trivialize this particular problem. Builtins for vector norms, etc. are allowed though.

Hint to get started:

Think about a circle around each of those 3 points with their distance to you as radius.

# Test cases

Input format for one point here is [[x,y],d] with x and y being the coordinates and d being the distance to this point. The 3 of those points are arranged in a list. Output will be x and then y in a list.

[[[1, 2], 1.414], [[1, 1], 2.236], [[2, 2], 1.0]] -> [2, 3]
[[[24.234, -13.902], 31.46], [[12.3242, 234.12], 229.953], [[23.983, 0.321], 25.572]] -> [-1.234, 4.567]
[[[973.23, -123.221], 1398.016], [[-12.123, -98.001], 990.537], [[-176.92, 0], 912.087]] -> [12.345, 892.234]


You can generate additional test cases with this Pyth program. Location goes on the first line of the input and the 3 points are on the following 3 lines.

Happy Coding!

• Because it needs to fit onto a two-dimensional Cartesian plane, the code needs to be as short as possible. Commented Mar 20, 2016 at 8:31
• You are obviously using inexact results which still can result in ambiguity, how should we handle this? Commented Mar 20, 2016 at 13:16
• @flawr Just assume that all results are exact und unique. Your program should work for cases with very few decimal places, don't worry about ambiguity. I will clean the challenge up when I'm home. Commented Mar 20, 2016 at 14:09
• Since the accepted answer is effectively just a graphing calculator, I'll mention that there's a 96-byte solution in TI-Basic (68k version). To clarify, is solve (given three circle equations) trivializing the problem? I thought it was, but if you're okay with such things I'll go ahead and post it.
– Fox
Commented Mar 27, 2016 at 18:46

# Desmos, 122 bytes

Online use. Copy+paste each equation into an equation box, click "add all" for each box, then click on the point of intersection, then enter in each value as appropriate. each of A, B, and C are the distances for the points (a,b), (c,d), and (E,f), respectively. To get a square root in the value, type sqrt then the value in the box.

\left(x-a\right)^2+\left(y-b\right)^2=AA
\left(x-c\right)^2+\left(y-d\right)^2=BB
\left(x-E\right)^2+\left(y-f\right)^2=CC


Or you can take a look here:

• This looks pretty nice, but does this satisfy our criteria for programming languages? This looks just like a plotting tool for me, never used it tho, so I might be wrong. Commented Mar 19, 2016 at 0:30
• @DenkerAffe I actually asked a question about it on meta before, and it does indeed count. Commented Mar 19, 2016 at 0:33
• Alright, have my upvote then :) Commented Mar 19, 2016 at 0:36
• I dispute the byte count. You get clicking the point of intersection for free, which doesn't seem right. Commented Mar 21, 2016 at 22:04
• @lirtosiast I would argue that the point is always there by definition and further state that the OP said that this sort of interaction was okay. If, however, you believe there should be some penalty, I'm open to suggestions. Commented Mar 21, 2016 at 22:10

## C, 362348 345 bytes

Input is given as a sequence of space-separated floats on stdin: x1 y1 d1 x2 y2 d2 x3 y3 d3. Output is similar on stdout: x y.

#define F"%f "
#define G float
#define T(x)(b.x*b.x-a.x*a.x)
typedef struct{G a;G b;G c;}C;G f(C a,C b,G*c){G x=b.b-a.b;*c=(T(a)+T(b)-T(c))/x/2;return(a.a-b.a)/x;}main(){C a,b,c;G x,y,z,t,m;scanf(F F F F F F F F F,&a.a,&a.b,&a.c,&b.a,&b.b,&b.c,&c.a,&c.b,&c.c);x=f(a,a.b==b.b?c:b,&y);z=f(b.b==c.b?a:b,c,&t);m=t-y;m/=x-z;printf(F F"\n",m,x*m+y);}


C is a structure type whose members are an x-coordinate a, a y-coordinate b, and a distance (radius) c. The function f takes two C structures and a pointer to a float, and determines the line at which the C (circles) intersect. The y-intercept of this line is placed into the pointed-to float, and the slope is returned.

The program calls f on two pairs of circles, then determines the intersection of the produced lines.

• Since we've clarified that the inputs are not collinear, the lines f produces will never be parallel. The tests are to ensure they are also not vertical, since I'm using slope-intercept form. This way, there is always exactly one answer.
– Fox
Commented Mar 20, 2016 at 7:39
• 329 bytes Commented Sep 5, 2020 at 23:48

# APL (Dyalog Unicode) 18.0, 23 bytes

-∘(+/)⍥(×⍨)⌹⍥(2-⌿⊢)¯2×⊢


Try it online!

A tacit infix function that can be called like dists f pts, where pts contains three centers' coordinates as a 3-row 2-column matrix, and dists contains three distances to the respective centers.

Uses 18.0 feature ⍥. TIO's Dyalog Unicode is not updated yet from 17.1, so TIO link uses Dyalog Extended for demonstration purposes.

### How it works

Uses the same method as R.T.'s Python answer, i.e. constructing a linear system of equations and solving it.

The distance equations are necessarily quadratic. Then how did we get linear ones?

Let's consider just two circles, and denote the point we want to find as $$\P\$$, two circles' centers as $$\C, D\$$, and their radii as $$\r_C, r_D\$$. Also, let's use $$\A_x, A_y\$$ for the $$\x\$$- and $$\y\$$-coordinates of a point $$\A\$$. Then the following equations hold:

$$(P_x-C_x)^2+(P_y-C_y)^2 =r_C^2 \\ (P_x-D_x)^2+(P_y-D_y)^2 =r_D^2 \\$$

Then we expand them and subtract the second from the first:

$$P_x^2-2P_xC_x+C_x^2+P_y^2-2P_yC_y+C_y^2=r_C^2\\ P_x^2-2P_xD_x+D_x^2+P_y^2-2P_yD_y+D_y^2=r_D^2\\ -2P_x(C_x-D_x)+(C_x^2-D_x^2)-2P_y(C_y-D_y)+(C_y^2-D_y^2)=r_C^2-r_D^2\\ -2P_x(C_x-D_x)-2P_y(C_y-D_y)=(r_C^2-C_x^2-C_y^2)-(r_D^2-D_x^2-D_y^2)$$

We've just got one linear equation on $$\P_x\$$ and $$\P_y\$$. Since we have three circles given, we can take any two pairs and construct two equations, which then solving the system of equations gives the coordinates we want.

-∘(+/)⍥(×⍨)⌹⍥(2-⌿⊢)¯2×⊢  ⍝ Left: dists, Right: pts
-∘(+/)⍥(×⍨) ⍥(2-⌿⊢)      ⍝ Constants part
⍥(×⍨)              ⍝ Square each number in both dists and pts
∘(+/)                   ⍝ Row sums of squared pts
-                        ⍝ Subtract above from squared dists
⍥(2-⌿⊢)      ⍝ Take pairwise difference

⍥(2-⌿⊢)¯2×⊢  ⍝ Coefficients part
¯2×⊢  ⍝ -2 times pts
⍥(2-⌿⊢)      ⍝ Take pairwise difference between rows
⌹             ⍝ Solve linear equations


# Python - 172

Takes input as a list of tuples of form (x,y,d). Let me know if you see a way to golf this further, I feel like there must be but I can't figure it out!

import numpy as N
def L(P):
Z=[p[2]**2-p[0]**2-p[1]**2 for p in P];return N.linalg.solve([[P[i][0]-P[0][0],P[i][1]-P[0][1]]for i in[1,2]],[(Z[0]-Z[i])*0.5 for i in [1,2]])

• You can omit some whitespaces here. Something like -1 if 1 else 1 can become -1if 1else 1 for example. This also works with brackets.There are a few places where you can take advantage of that. Also .5 is the same as 0.5. Commented Mar 22, 2016 at 13:57