Coordinates in a circle from 3 distances

Question

Suppose I have a circle with a radius $r$ with coordinate system whose origin is at the center. Suppose we have a point with unknown coordinates $(x,y)$ in our coordinate system, with three known distances to the edge of the circle $d_1$, $d_2$, and $d_3$, where $d_1$ is 0 degrees (positive y), $d_2$ is pointed at 120 degrees, and $d_3$ is pointed at 240 degrees. Using $r$, $d_1$, $d_2$, and $d_3$ calculate the coordinates $(x,y)$ for the point.

Edit: As Emilio points out in his answer the question may be over constrained (there was a slight misunderstanding on the question being asked). However what I am really looking for is the best approximation (it is a real world problem using sensors which are likely noisy), so the assumption can be made that the given the first two distances the third will be pretty close to working out. A solution involving taking the average of three coordinates obtained by using pairs of distances would be sufficient.

Answer 1

This problem amounts to finding the center of a circle that passes through three non-colinear points. The coordinates you’re looking for will be the negation of the coordinates of this center point.

There are several ways to do this, but since we’re only interested in finding the center, the method of bisectors seems best.

The center of a circle lies at the intersection of the perpendicular bisectors of its chords. So, once we have the equations of two chord bisectors, this problem reduces to solving a system of linear equations. Our three points are $$\begin{align}P_1&=d_1(0,1)\\P_2&=d_2\left(\frac{\sqrt3}2,-\frac12\right)\\P_3&=d_3\left(-\frac{\sqrt3}2,-\frac12\right)\end{align}.$$ To simplify the calculations, we’ll take the segments $\overline{P_1P_2}$ and $\overline{P_1P_3}$ as our two chords. The equations of the perpendicular bisectors are then $$\begin{align}(P_2-P_1)\cdot\left(P-{P_2+P_1\over2}\right)&=0\\(P_3-P_1)\cdot\left(P-{P_3+P_1\over2}\right)&=0\end{align}$$ or, after a bit of rearrangement $$\begin{align} \sqrt3d_2x-(2d_1+d_2)y&=d_2^2-d_1^2 \\ \sqrt3d_3x+(2d_1+d_3)y&=d_1^2-d_3^2. \end{align}$$ You can take it from here, I’m sure. As a sanity check, you can compute the lengths of $\overline{CP_1}$, $\overline{CP_2}$ and $\overline{CP_3}$ and check that they’re close to (ideally, equal to) the known radius $r$.

Answer 2

In general the problem is overdetermined and have no solutions.
Look at the figure: $B$,$B'$ and $B''$ are the three points on the circle and $C$ is the point that we want to find.

If It has a fixed distance ( say $d_1$) from $B$ than it stay on the circle centerd at $B$ and with radius $d_1$ ( the red circle in the figure).

If It has a fixed distance ( say $d_2$) from $B'$ than it stay on the circle centerd at $B'$ and with radius $d_2$ ( the blue circle in the figure).

These two circles have at most two common points, and the distance of these points from $B''$ cannot be chosen freely .

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I've added another figure that seems better for the problem.

Using the ditances from the points $B$ and $B'$ we can find the coordinates of $C$ and $F$ as intersection of the two circle. Now you can compare the measured distance from $B''$ with these two points. If the real measures are correct, one of the points $C$ $E$ has a distance frm $B''$ that is near the measured distance, and we use some best fit method to refine the solution.