356. Find the center and radius of the sphere

$$
x^{2}+y^{2}+z^{2}-2 x+6 y-6=0
$$
Solution. We will transform the given equation into the form (1)

\[
\begin{gathered}
x^{2}+y^{2}+z^{2}-2 x+6 y-6=\left(x^{2}-2 x+1\right)+\left(y^{2}+6 y+9\right)+ \\
+z^{2}-1-9-6=(x-1)^{2}+(y+3)^{2}+z^{2}-16=0
\end{gathered}
\]

or

\[
(x-1)^{2}+(y+3)^{2}+z^{2}=16
\]

from which

\[
a=1, b=-3, c=0, R^{2}=16
\]

i.e., the center of the sphere is at point \( C(1, -3, 0) \) and its radius is \( R=4 \).