Answer
The sphere is centered at $C(-G / 2,-H / 2,-I / 2)$ and its radius is
\[
r=\sqrt{\frac{1}{4}\left(G^{2}+H^{2}+I^{2}\right)-J}=\frac{\sqrt{K}}{2}
\]
Work Step by Step
Let us consider the equation
\[
0=G x+H y+I z+J+z^{2}+y^{2}+x^{2}
\]
By completing squares:
\[
\begin{aligned}
-1= z^{2}+I z+y^{2}+H y+ x^{2}+G x\\
x^{2}+G x+\frac{G}{4}+y^{2}+H y+\frac{H^{2}}{4}+z^{2}+I z+\frac{I^{2}}{4}=\frac{1}{4}\left(G^{2}+H^{2}+I^{2}\right)-J \\
\Rightarrow\left(x+\frac{G}{2}\right)^{2}+\left(y+\frac{H}{2}\right)^{2}+\left(z+\frac{I}{2}\right)^{2}=\frac{1}{4}\left(G^{2}+H^{2}+I^{2}\right)-J &
\end{aligned}
\]
If $\frac{K}{4}=\frac{1}{4}\left(G^{2}+H^{2}+I^{2}\right)-J>0,$ then the equation corresponds to a sphere of radius $r=\sqrt{K} / 2$ centered at $C(-G / 2,-H / 2,-I / 2)$
If $K / 4=0 \Rightarrow K=0$
and then
\[
\begin{array}{c}
\left(\frac{I}{2}+z\right)=\left(\frac{G}{2}+x\right)=\left(\frac{H}{2}+y\right)=0 \\
\Rightarrow \quad y=-\frac{H}{2}, \quad x=-\frac{G}{2} \quad \text { and } \quad z=-\frac{I}{2}
\end{array}
\]
These are the coordinates of a point in 3-space.
Since $x^{2} \geq 0$ for all real $x$, $4 K \geq 0$. Therefore, if $K<0$, the equation doesn't have a solution. This means that the equation has no graph.
b) The sphere is centered at $C(-G / 2,-H / 2,-I / 2)$ and the radius is
\[
\sqrt{\frac{1}{4}\left(I^{2}+H^{2}+G^{2}\right)-J}=\frac{\sqrt{K}}{2}=r
\]
