How would I go about calculating the center of a sphere based on points from it's surface?

Question

I've recorded random Cartesian coordinates from the surface of a sphere. I'm trying to extrapolate where the center x,y,z coordinate of this sphere would be based on the points below.

I'm trying to create a script to solve this, but the math has me stuck.

Appreciate any help. Thanks!

Coordinates x,y,z:
-1.190185,0.7824033,0.1691585
-1.152931,0.7811859,0.1401751
-1.110813,0.775885, 0.1348239
-1.078485,0.772419, 0.1545465
-1.101839,0.7839018,0.2090827
-1.120642,0.7859643,0.2231803
-1.136541,0.7895643,0.2040433
-1.135001,0.7879777,0.1715891
-1.111366,0.7854245,0.1738369
-1.115193,0.78763 , 0.1931185
-1.130671,0.7892017,0.1985596
-1.132487,0.7872546,0.2240417
-1.161051,0.7876478,0.171136
-1.114549,0.7809206,0.1476536
-1.083838,0.7764122,0.1626304
-1.06429, 0.769419,0.2177327
-1.142459,0.7805369,0.2528498
-1.162153,0.7863876,0.1668259

Answer 1

The equation of a sphere with center $(x_0, y_0, z_0)$ and radius $r$ is

$ (x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 = r^2 $

Expanding, we get,

$ x^2 + y^2 + z^2 + A x + B y + C z + D = 0 $

where

$ A = -2 x_0 , \ B = - 2 y_0, \ C = - 2 z_0, \ D = x_0^2 + y_0^2 + z_0^2 - r^2 $

To identify the parameters $A,B,C,D$, we build a linear regression model, where we set

$ A x_i + B y_i + C z_i + D = -( x_i^2 + y_i^2 + z_i^2) $

If the above model equation is repeated for $i = 1, 2,.., N $, then we end up with the following regression model

$ M X = Y $

where the $i$-th row of $M$ is

$M_i = [x_i , y_i, z_i, 1]$

and $X$ is the vector of parameters:

$ X = [A, B, C, D]^T$

and $Y$ is the data vector, whose $i$-th entry is

$Y_i = -( x_i^2 + y_i^2 + z_i^2) $

The least squares estimate of $X$ is given by

$ \hat{X} = (M^T M)^{-1} M^T Y $

The computation of $\hat{X}$ involves the inversion of the $4 \times 4$ matrix $M^T M$.

Once $\hat{X}$ is determined, then we know the estimate of $A, B, C, D$, and hence,

$x_0 = - \dfrac{1}{2} A$

$y_0 = - \dfrac{1}{2} B $

$z_0 = -\dfrac{1}{2} C $

$ r^2 = x_0^2 + y_0^2 + z_0^2 - D$

Answer 2

One way to do it is the following:

Take any three distinct points $A,B,C,$ and find the circumcentre of those $3$ points. To do this, just find the perpendicular bisector of $AB$ and then find the perpendicular bisector of $BC.$ The circumcentre of triangle $ABC$ is the intersection of these two perpendicular bisectors.

Then the centre of the sphere lies of the line perpendicular to the plane formed by the three points and that passes through the circumcentre of $ABC.$

Now do this again with three other points $DEF$: and where the line perpendicular to triangle $ABC$ and the line perpendicular to triangle $DEF$ intersect is the centre of the sphere.

Answer 3

The equation of a 3D sphere is $$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=r^2$$, where $(x_0, y_0, z_0)$ is the center of the sphere and $r$ is the radius. Plug in any four noncoplanar points $(x,y,z)$ to get a system of equations with four unknowns $x_0, y_0, z_0, r$ in order to solve for $z$. Of course, this system is quadratic, so it would be tricky to solve by hand, but it can be solved exactly either by means of an existing computer program or approximated through numerical means.