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How to prove that there is always one non-trivial real solution $(x_0,y_0,z_0,r)$ for the following simultaneous equations?

$$ (x_i-x_0)^2 + (y_i-y_0)^2 + (z_i-z_0)^2 = r^2 $$

for $i=1,2,3,4$, and

$$\left| \begin{array}{ccc} x_2-x_1 & x_3-x_1 & x_4-x_1 \\ y_2-y_1 & y_3-y_1 & y_4-y_1 \\ z_2-z_1 & z_3-z_1 & z_4-z_1 \end{array} \right|\not= 0 $$

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  • $\begingroup$ What would be a trivial solution? $\endgroup$ – Rodrigo de Azevedo Jul 3 '16 at 2:21
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It seems that you're going to prove existence of circumcenter of non-degenerated tetrahedron (i.e not all the vertices are on same plane, which is exactly the determinant condition in your question). We need a lemma :

Let $S$ be a plane in $\mathbb{R}^{3}$ with normal vector $\mathbf{u}$ and $l$ be a line with direction vector $\mathbf{v}$. If $\mathbf{u}$ and $\mathbf{v}$ are not orthogonal, then there exists (unique) point $P\in l\cap S$.

Proof will be easy, maybe (If you imagine the situation, it is intuitively clear). Using this, we can prove the existence of circumcenter. Consider a plane $H$ containing $P_{1},P_{2},P_{3}$ and let $Q$ be a circumcenter of triangle $P_{1}P_{2}P_{3}$ that lies on $H$. Let $l$ be a line passes $Q$ with direction vector parallel to normal vector of $H$. Let $M$ be a midpoint of $P_{1}$ and $P_{4}$ and consider a plane $S$ contains $M$ and has a normal vector parallel to $P_{1}-P_{4}$. Then we can check this $l,S$ satisfy the condition of lemma and the point $C\in l\cap S$ became circumcenter.

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  • $\begingroup$ I need an analytic approach using Linear Algebra $\endgroup$ – Well Harassed Programmer Jul 3 '16 at 14:45
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The simultaneous equations can be written as,

$$\begin{cases} (x_1-x_0)^2+(y_1-y_0)^2+(z_1-z_0)^2=r^2 ~~~~~~~~~(1)\\ (x_2-x_0)^2+(y_2-y_0)^2+(z_2-z_0)^2=r^2 ~~~~~~~~~(2)\\ (x_3-x_0)^2+(y_3-y_0)^2+(z_3-z_0)^2=r^2 ~~~~~~~~~(3)\\ (x_4-x_0)^2+(y_4-y_0)^2+(z_4-z_0)^2=r^2 ~~~~~~~~~(4) \end{cases}$$

(4) subtract by (1), (3) subtract by (1) and (2) subtract by (1), we get,

$$\begin{cases} (x_4-x_1)(x_4+x_1-2x_0)+(y_4-y_1)(y_4+y_1-2y_0)+(z_4-z_1)(z_4+z_1-2z_0)=0 ~~~~~~~~~(1)\\ (x_3-x_1)(x_3+x_1-2x_0)+(y_3-y_1)(y_3+y_1-2y_0)+(z_3-z_1)(z_3+z_1-2z_0)=0 ~~~~~~~~~(1)\\ (x_2-x_1)(x_2+x_1-2x_0)+(y_2-y_1)(y_2+y_1-2y_0)+(z_2-z_1)(z_2+z_1-2z_0)=0 ~~~~~~~~~(1)\\ \end{cases}$$

Because the following determinant isn't zeor, there exists an unique solution. The proof is complete.

$$\left| \begin{array}{ccc} x_2-x_1 & x_3-x_1 & x_4-x_1 \\ y_2-y_1 & y_3-y_1 & y_4-y_1 \\ z_2-z_1 & z_3-z_1 & z_4-z_1 \end{array} \right|$$

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