Finding $x, y$ coordinates in a circle I have a laser that will be measuring distances in a circular tank to identify  unique locations. The laser will take north $(y)$ and west $(x)$ measurements and then it can be rotated if necessary to take south and east measurements. The angle of the two measurements will be $90$ degrees. The diameter of the tank is known.
How can I find the coordinates of the laser from the measurements it takes? 

Thanks in advance.
 A: Taking only two anti-parallel measurements, (N and S) or (E and W) is not sufficient to resolve any ambiguity. However, taking two orthogonal measurements is sufficient, if the calculation is done as follows:
Recognize that a distance measurement $N$ in the y direction provides enough information to determine that the point $(x,y)$ is on a circle of radius $r$ defined by the curve $x^2+(y-N)^2 = r^2$. You also know that the point $(x,y)$ must lie within the original circle $x^2+y^2 = r^2$. That limits the solution to an arc that is convex up.
Also, a distance measurement $E$ in the x direction provides enough information to determine that the point $(x,y)$ is on a circle of radius $r$ defined by the curve $(x-E)^2+y^2 = r^2$. You also know that the point $(x,y)$ must lie within the original circle $x^2+y^2 = r^2$. That also limits the solution to another arc  that is convex to the right.
The solution $(x,y)$ is where these two arcs intersect. It satisfies the condition that it is at the intersection of the two offset circles AND it must be inside the original circle. (The second ambiguous point of intersection for the offset circle will lie outside the original circle.)
I'll leave the math to you. It does require solving a quadratic equation AND determining which solution lies within the original circle.
A: I solved the problem and here's a picture of it.
I hope it's understandable.
EDIT:
I realized that you can solve this with only two distances but they have to be opposite(i.e. N and S)
If we know N and S this is the way to get the coordinates:
$$y_0=\frac{S-N}{2}, x_0=\sqrt{r^2-\frac{(N+S)^2}{4}}$$
A: Assume that the origin is the centre of the circle, and the west measurement is $x$ and the north measurement is $y$, and both are positive.
Interestingly, the problem is curiously intractable if you only have the $x$ and the $y$, as the resulting equations become fourth-order polynomial equations in $a$ and in $b$, where $(a,b)$ are the coordinates of the laser.
However if you also have the south and east measurements, let's call these distances $y'$ and $x'$, both positive quantities, then the problem becomes a lot simpler.
Pythagoras gives us $$(x-a)^2+b^2=R^2$$ and $$(x'+a)^2+b^2=R^2$$
These result in $$a=\frac {x-x'}{2}$$
Similarly, we get $$b=\frac {y'-y}{2}$$
A: I believe I have a way of doing it with only having two measurements. 


