Equation of the sphere that passes through 4 points Write he equation of the sphere that passes through points
$$a(-5,4,1),b(3,4,-5),c(0,0,4),d(0,0,0)$$
I tried to use four points to draw a geometric shape and then calculate the center of this shape on the basis of the circle that passing  on four points. But I did not succeed

Here is the book answer
  $$x^2+y^2+z^2+54x−58y+4z=0$$

 A: $$\left | \begin{matrix} x^2 + y^2 + z^2 & x & y & z & 1 \\
                         x_1^2 + y_1^2 + z_1^2 & x_1 & y_1 & z_1 & 1 \\
                         x^2_2 + y_2^2 + z_2^2 & x_2 & y_2 & z_2 & 1 \\
                         x_3^2 + y_3^2 + z_3^2 & x_3 & y_3 & z_3 & 1 \\
                         x_4^2 + y_4^2 + z_4^2 & x_4 & y_4 & z_4 & 1                          \end{matrix} \right |  = 0 $$
i.e. $-128(x^2+y^2+z^2-4x-\frac{29}{2}y-4z)=0$
A: Using the equation for points on spheres:  
$\qquad(x-a)^2+(y-b)^2+(z-c)^2=r^2$
Using coordinates of the four points provided, we have four simultaneous equations to solve for $a, b, c, d$.
\begin{cases}
(-5-a)^2+(4-b)^2+(1-c)^2=r^2  \\
(3-a)^2+(4-b)^2+(-5-c)^2=r^2  \\
(0-a)^2+(0-b)^2+(4-c)^2=r^2  \\
(0-a)^2+(0-b)^2+(0-c)^2=r^2 
\end{cases}
(Substracting third and fourth equation yield $c$, and the first two yield $a$ then $b$, then you have $r$, easy to solve on paper)  
and get a nonnegative solution:
\begin{cases}
a=2  \\
b=\frac{29}{4}  \\
c=2  \\
r=\frac{\sqrt{969}}{4}
\end{cases}
(just solved it, maybe you can check the answers :D )
A: The technique here is straightforward.
First take the general equation of a sphere:
$$(x-a)^2+(y-b)^2+(z-c)^2=r^2$$
If this passes through the points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ then you have two equations:
$$(x_1-a)^2+(y_1-b)^2+(z_1-c)^2=r^2$$$$(x_2-a)^2+(y_2-b)^2+(z_2-c)^2=r^2$$
Now subtract the second from the first $$x_1^2-x_2^2+2a(x_2-x_1)+y_1^2-y_2^2+2b(y_2-y_1)+z_1^2-z_2^2+2c(z_2-z_1)=0$$
You now have a linear equation in three unknowns, $a, b, c$ - the inconvenient square terms all cancel. You have enough data to obtain three independent equations, which you should be able to solve. Here you can do somewhat better by choosing the point $(0,0,0)$ as the second point in each case, which gives $$2ax_1+2by_1+2cz_1=x_1^2+y_1^2+z_1^2$$ and gives you some simpler arithmetic.
A: the equation of a sphere is given by $$(x-x_M)^2+(y-y_M)^2+(z-z_M)^2=R^2$$
inserting the given coordinates you will get
$$(-5-x_M)^2+(4-y_M)^2+(1-z_M)^2=R^2$$ (I)
$$(3-x_M)^2+(4-y_M)^2+(-5-z_M)^2=R^2$$ (II)
$$x_M^2+y_M^2+(4-z_M)^2=R^2$$ (III)
$$x_M^2+y_M^2+z_M^2=R^2$$ (IV)
solve this System for $x_M,y_M,z_M$ and $R$
and now subtract (I)-(II) for example
A: The equation of the sphere with a centre in $(x_0,y_0,z_0) $ and radius $r $ is: $(x-x_0)^2+(y-y^0)^2+(z-z_0)^2=r^2$. For us, the unknowns are $x_0,y_0,z_0,r $.
Plug the given four points in:
$$\begin {align}(-5-x_0)^2+(4-y_0)^2+(1-z_0)^2=r^2 \\ (3-x_0)^2+(4-y_0)^2+(-5-z_0)^2=r^2 \\ (-x_0)^2+(-y_0)^2+(4-z_0)^2=r^2 \\  (-x_0)^2+(-y_0)^2+(-z_0)^2=r^2\end {align} $$
or (after a bit of transforming):
$$\begin {align}42+10x_0-8y_0-2z_0+x_0^2+y_0^2+z_0^2=r^2 \\ 50-6x_0-8y_0+10z_0+x_0^2+y_0^2+z_0^2=r^2 \\ 16-8z_0+x_0^2+y_0^2+z_0^2=r^2 \\ x_0^2+y_0^2+z_0^2=r^2\end {align} $$
Now, take away one of the equations from the rest (in this case, the last one is the easiest one):
$$\begin {align}42+10x_0-8y_0-2z_0=0 \\ 50-6x_0-8y_0+10z_0=0\\ 16-8z_0=0\\ x_0^2+y_0^2+z_0^2=r^2\end {align} $$
Solve the system of the first three (linear) equations, to obtain: $x_0=2, y_0=\frac {29}{4}, z_0=2$. Plug that into the fourth to obtain $r=\sqrt {2^2+\left (\frac {29}{4}\right)^2+2^2}=\frac {\sqrt {969}}{4}$.
Thus, the final equation is:
$$(x-2)^2+(y-\frac {29}{4})^2+(z-2)^2=\frac {969}{16} $$
or, after transforming:
$$x^2+y^2+z^2-4x-\frac {29}{2}y-4z=0$$
as all constant terms will happen to cancel each other.
A: Here is a general and closed solution. Call the four non-coplanar points $A$, $B$, $C$ and $D$. The center of the sphere is where the planes between $B$ and $A$, $C$ and $A$, and $D$ and $A$ intersect. For instance, the plane between $B$ and $A$ has the normal $B-A$, and goes through $(A+B)/2$, i.e. has the equation
$$ (B-A)\cdot I=(B-A)\cdot(B+A)/2\;.$$
It helps a bit to work in coordinates in which the origin is at $A$, so that $B'=B-A$, $C'=C-A$ and $D'=D-A$. In those coordinates, $A'=0$ and thus the plane equations are of the form
$$ B'\cdot I=|B'|^2/2\;,$$
and likewise for $C'$ and $D'$. We can either write this as a matrix equation,
$$ N\,I:=(B',C',D')^T\,I=\frac{1}{2}(|B'|^2,|C'|^2,|D'|^2)^T\;,$$
or just solve this a bit more "geometrically", the results are of course the same. What is meant by "geometrical" solution is that, since the points are not coplanar, we can make the ansatz
$$ I=\beta\,C'\times D'+\gamma\,D'\times B'+\delta\,B'\times C' \;.$$
Obviously, $C'\cdot (C'\times D')=0$ and so on, and the only nonvanishing combinations are $B'\cdot(C'\times D')$ and cyclic permutations thereof, which all equal to $\det N$. (If and only if the points were all on a common plane, $\det N=0$, and there was no unique sphere.) This determines
$$ \beta\,\det N=|B'|^2/2$$
etc. So in the shifted coordinate system the center of the sphere is at
$$ I'=\frac{1}{2B'\cdot(C'\times D')}\left(|B'|^2\,C'\times D'
+ |C'|^2\,D'\times B'+ |D'|^2\,B'\times C'\right)\;.$$
In the original, unshifted coordinates
$$ I=A+I'\;.$$
Given the center, the computation of the radius is trivial, $r=\sqrt{|I'|^2}$.
