# Finding the fourth vertex of a tetrahedron bounded by 4 spheres

I am trying to solve for the 4th vertex of a tetrahedron with 4 known vertices. The vertices of the tetrahedron are the centers of 4 spheres. Basically I need to find the coordinates of the 4th sphere whose radius is also already known. A set of formulas are actually given to solve this but I can't seem to figure out how to use them: $$(x_d-x_a)^2 + (y_d-y_a)^2 + (z_d-z_a)^2=(r_d+r_a)^2$$ $$(x_d-x_b)^2 + (y_d-y_b)^2 + (z_d-z_b)^2=(r_d+r_b)^2$$ $$(x_d-x_c)^2 + (y_d-y_c)^2 + (z_d-z_c)^2=(r_d+r_c)^2$$ a, b, and c represent the circles of known positions and the d is the one whose x,y,z coordinates need to be determined. Again, all radii of all 4 spheres are known.

The system would look like this: sphere bounded tetrahedron

Thank you very much in advance!

• You have a system of three equations with 3 unknowns. That system of equations expresses the condition that your fourth vertex $d$ lies at the intersection of three spheres, so it will generally have two distinct solutions. – N.Bach Jun 21 '18 at 10:30

Subtract the equations from each other to eliminate the squares. That leaves you with two linearly independent linear equations for the three unknowns that you can use to express any one of them as a linear function of any other one of them. For instance, express $y_d$ and $z_d$ as linear functions of $x_d$. Substitute into one of the equations, solve the resulting quadratic equation for $x_d$ and substitute the result into $y_d$ and $z_d$ as functions of $x_d$.