Equation of sphere circumscribing the tetrahedron bounded by planes $x+y=0$, $y+z=0$, $z+x=0$, $x+y+z=1$ I found a following question in 1st year undergrad course:

Find the equation of sphere circumscribing the tetrahedron bounded by the planes 
  $$x+y=0 \qquad y+z=0 \qquad z+x=0 \qquad x+y+z=1$$

Can anyone tell me in detail how to find the equation of sphere ?
I know if i get four vertices of tetrahedron then i can able find out the equation. But how to find the vertices of tetrahedron?
 A: There are $4$ planes
$$\begin{cases} 
P_0 &: x + y + z = 1\\
P_1 &: y + z = 0\\
P_2 &: x + z = 0\\
P_3 &: x + y = 0
\end{cases}$$
To find the vertices, select $3$ planes from $P_0,P_1,P_2,P_3$ and intersect them. 
Each vertex corresponds to a different way to select the planes and there are totally
$4$ selections. For $i = 0, \ldots, 3$, let $v_i = (x_i,y_i,z_i)$ be the vertex belongs to $\bigcap\limits_{j=0,\ne i}^3 P_j$.
For $v_0 \in P_1\cap P_2 \cap P_3$, it is clear $v_0 = (0,0,0)$.
For $v_1 \in P_0\cap P_2 \cap P_3$, since it belongs to $P_2 \cap P_3$, we have
$$x_1 + z_1 = 0, x_1 + y_1 = 0 \implies v_1 = (x_1, -x_1, -x_1)$$
Substitute this into the equation of $P_0$, we get $x_1 = -1 \implies v_1 = (-1,1,1)$
By a similar manner, we can determine the coordinates of $v_2, v_3$. The end result is
$$\begin{cases}
v_0 = (0,0,0)\\
v_1 = (-1,1,1)\\
v_2 = (1,-1,1)\\
v_3 = (1,1,-1)
\end{cases}$$
Given coordinates of the four vertices, the circumsphere is given by following equation:
$$\left|\begin{matrix}
1 & x & y & z & x^2 + y^2 + z^2\\
1 & x_0 & y_0 & z_0 & x_0^2 + y_0^2 + z_0^2\\
1 & x_1 & y_1 & z_1 & x_1^2 + y_1^2 + z_1^2\\
1 & x_2 & y_2 & z_2 & x_2^2 + y_2^2 + z_2^2\\
1 & x_3 & y_3 & z_3 & x_3^2 + y_3^2 + z_3^2\\
\end{matrix}\right|
= 0$$
Using the coordinates of $v_i$ derived above, we get
$$
\left|\begin{matrix}
1 & x & y & z & x^2 + y^2 + z^2\\
1 & 0 & 0 & 0 & 0\\
1 & -1 & 1 & 1 & 3\\
1 &  1 & -1 & 1 & 3\\
1 & 1 & 1 & -1 & 3\\
\end{matrix}\right| 
= -
\left|\begin{matrix}
 x & y & z & x^2 + y^2 + z^2\\
-1 & 1 & 1 & 3\\
 1 & -1 & 1 & 3\\
 1 & 1 & -1 & 3\\
\end{matrix}\right| = 0$$
This can be further simplified to
$$\bbox[border:1px solid blue;padding: 1em;]{x^2+y^2+z^2 - 3(x+y+z) = 0}$$
If you don't like evaluating a $5 \times 5$ determinant, here is another
way to determine the circumsphere. 
Just like circle inversion in the plane, if you perform a sphere inversion with respect to the unit sphere in $\mathbb{R}^3$, any sphere passing through the origin will get mapped to a plane. Since one of the vertex $v_0$ is origin, the circumsphere
we seek will get mapped to a plane.
Under the sphere inversion, $v_1, v_2, v_3$ get mapped to $\frac13 v_1$, $\frac13 v_2$ and $\frac13 v_3$. It is easy to see they are lying on the plane $x + y + z = \frac13$. If we reverse the sphere inversion, we obtain following equation for the circumsphere.
$$\frac{x+y+z}{x^2+y^2+z^2} = \frac13 \iff (x^2+y^2+z^2) - 3(x+y+z) = 0$$
This is the same equation we obtained before using a $5\times 5$ determinant.
A: Perhaps not the fastest, but a very straight forward method is the following:


*

*Find the four vertices of the tetrahedron

*Write the sphere equation as $x^2+y^2+z^2+Ax+By+Cz+D=0$

*Substitute the four vertices in the equation to get a $4\times4$ system.

*Solve it for $A$, $B$, $C$ and $D$.

