I have $6$ equations and 6 unknowns, but I have trouble solving them. The problem I have is when I start substituting them, I get a quadratic formula in a square root that I can't simplify and reduce further.
The unknowns are $x, y, z, m, n, o$. (yes these are Cartesian coordinates), and the equations are:
- $0=(x−a)^2+(y−b)^2+(z−c)^2−s^2$
- $0=(x−d)^2+(y−e)^2+(z−f)^2−r^2$
- $0=(m−a)^2+(n−b)^2+(o−c)^2−h^2$
- $0=(m−d)^2+(n−e)^2+(o−f)^2−q^2$
- $0=(m−u)^2+(n−v)^2+(o−w)^2−t^2$
- $0=(x−m)^2+(y−n)^2+(z−o)^2−p^2$
Can anyone offer some guidance?
Thanks!
The equations $3,4,5$ can be solved independently. They are equations of spheres. Subtracting them in pairs, you get equations of planes. These are the planes of support of the intersection of the corresponding spheres (circles). Take the intersection of two planes to get a straight line. That straight line meets the three spheres in the same two points. (The resulting equation is quadratic).
Knowing $m,n,o$, the equations $1,2,6$ are of the same nature (three intersecting spheres).
In total, you will find four solutions.
A vector solution:
Assume the three spheres to be centered on $P,Q,R$, with radii $p,q,r$. We translate the three points to be $0,Q':=Q-P,R':=R-P$. Let the intersection point be $X$. We have
$$\begin{cases}(X-0)^2&=p^2,\\(X-Q')^2&=q^2,\\(X-R')^2&=r^2.\end{cases}$$
Eliminating $X^2$,
$$\begin{cases}2Q'X=Q'^2+p^2-q^2&,\\2R'X=R'^2+p^2-r^2&.\end{cases}$$
This is a linear system of two equations in three unknowns. Compute two distinct solutions, let $A$ and $B$, and the general solution is $X=A+tB'$ where $B':=B-A$.
Now, $$X^2=(A+tB')^2=p^2$$ is a quadratic equation in $t$ that gives the two solutions. Translate back by $P$.