Find the sphere through the vertices of the tetrahedron…

Find the sphere through the vertices of the tetrahedron bounded by the plane $x+y+z=a$ and the coordinate planes.

Explanation of Method

The sphere through the four vertices of the tetrahedron cut off by a plane from the coordinate trihedron.

Step 1: Vertices

\[ O(0,0,0),\quad A(a,0,0),\quad B(0,a,0),\quad C(0,0,a). \]

Step 2: Coefficients

Write the sphere as $x^2+y^2+z^2+Dx+Ey+Fz+G=0$. Then $O$ gives $G=0$. $A$ gives $a^2+Da=0$, so $D=-a$. In the same way $E=F=-a$.

\[x^2+y^2+z^2-a(x+y+z)=0\]

Step 3: Center and radius

\[ \left(x-\tfrac a2\right)^2+\left(y-\tfrac a2\right)^2+\left(z-\tfrac a2\right)^2=\tfrac{3a^2}4,\qquad S\left(\tfrac a2,\tfrac a2,\tfrac a2\right),\ \ r=\tfrac{a\sqrt3}2 . \]

Another way to see this: $O,A,B,C$ are four vertices of the cube with edge $a$ at the origin. The sphere is the cube's circumscribed sphere. Its center is the center of the cube, and its radius is half the space diagonal.

$O$ $A$ $B$ $C$ $S\left(\tfrac a2,\tfrac a2,\tfrac a2\right)$

The tetrahedron $OABC$ (for $a=2$) and its circumscribed sphere, center $S\left(\tfrac a2,\tfrac a2,\tfrac a2\right)$, radius $\tfrac{a\sqrt3}2$. $S$ lies on the diagonal $OS$ but not on the plane $ABC$.

Result

\[\boxed{\,x^2+y^2+z^2-a(x+y+z)=0\,}\]

Remark

Correction: the source says the center "lies on the plane $x+y+z=a$". It does not: $\tfrac a2+\tfrac a2+\tfrac a2=\tfrac{3a}2\neq a$. The center lies on the diagonal $x=y=z$, beyond the face $ABC$. The center of the circle through $A,B,C$ is $\left(\tfrac a3,\tfrac a3,\tfrac a3\right)$.