Find the center and radius of the sphere: 1^°…

Find the center and radius of the sphere: $1^\circ\ \ x^2+y^2+z^2-2x+2y-2z=0$ $2^\circ\ \ x^2+y^2+z^2+ax+by+cz+d=0$.

Explanation of Method

Reading off the center and radius of a sphere by completing squares.

Step 1: $1^\circ$

\[ (x-1)^2+(y+1)^2+(z-1)^2=1+1+1=3 . \]

\[C(1,-1,1),\qquad r=\sqrt3\]

There is no constant term, so $O$ lies on the sphere. Indeed $|CO|=\sqrt{1+1+1}=\sqrt3$.

Step 2: $2^\circ$

\[ \left(x+\tfrac a2\right)^2+\left(y+\tfrac b2\right)^2+\left(z+\tfrac c2\right)^2=\frac{a^2+b^2+c^2-4d}{4}. \]

\[C\left(-\tfrac a2,-\tfrac b2,-\tfrac c2\right),\qquad r=\tfrac12\sqrt{a^2+b^2+c^2-4d}\]

This is a real sphere only if $a^2+b^2+c^2-4d\gt 0$. If the expression is $0$, the equation describes the single point $C$. If it is negative, no real point satisfies it.

$C(1,-1,1)$ $O$ $\sqrt3$

$1^\circ$: center $C(1,-1,1)$, radius $\sqrt3$. The equation has no constant term, so the sphere passes through $O$, and $|CO|=\sqrt3$.

Result

\[\boxed{\,C(1,-1,1),\qquad r=\sqrt3\,}\]

\[\boxed{\,C\left(-\tfrac a2,-\tfrac b2,-\tfrac c2\right),\qquad r=\tfrac12\sqrt{a^2+b^2+c^2-4d}\,}\]