Find the radical centre of the four spheres :$x^2+y^2+z^2+2x+2y+2z+2=0,x^2+y^2+z^2+4y=0,x^2+y^2+z^2+3x-2y+8z+6=0,x^2+y^2+z^2-x+4y-6z-2=0$ Determine the radical centre of the spheres $x^2+y^2+z^2+2x+2y+2z+2=0,x^2+y^2+z^2+4y=0,x^2+y^2+z^2+3x-2y+8z+6=0,x^2+y^2+z^2-x+4y-6z-2=0.$
I tried solving the problem by assuming $S_1=x^2+y^2+z^2+2x+2y+2z+2,S_2=x^2+y^2+z^2+4y,S_3=x^2+y^2+z^2+3x-2y+8z+6,S_4=x^2+y^2+z^2-x+4y-6z-2.$ Now, the  radical plane of the $3$ spheres $S_1,S_2,S_3$ are $S_1-S_2=0,S_2-S_3=0,S_3-S_1=0.$ Now, these radical planes, intersect at the straight line $S_1=S_2=S_3.$  Thus, the radical axis of $S_1,S_2,S_3$ is $S_1=S_2=S_3.$ Similarly, the other possible radical axis of the spheres taken three at a time are: $S_1=S_2=S_4,S_1=S_3=S_4,S_2=S_3=S_4.$ Thus, the $4$ radical axis are $S_1=S_2=S_3,$$S_1=S_2=S_4,S_1=S_3=S_4,S_2=S_3=S_4.$ Now, the point where these $4$ radical lines intersects is called the radical centre. But how to calculate $C$ ? If we try to solve these $4$ equations, then I think we should require a fast processing and computing software or otherwise it becomes a tremendously huge calculation. Is there any shorter way to determine the radical centre $C$ using only elementary analytic geometry of spheres ?

Here is the computation.
$\quad\quad$ $(x,y,z)$ is the radical centre
$\iff S_1(x,y,z)=S_2(x,y,z)=S_3(x,y,z)=S_4(x,y,z)$
$\iff 2x+2y+2z+2=4y=3x-2y+8z+6=-x+4y-6z-2$
$\iff x=y-z-1\text{ and }x=2y-\frac83z-2=0\text{ and } x=-6z-2$
$\iff x=y-z-1\text{ and }y-\frac53z-1=0\text{ and } y+5z+1=0$
$\iff x=-\frac15, y=\frac12, z=-\frac3{10}$
I can hardly see why "we should require a fast processing and computing software or otherwise it becomes a tremendously huge calculation".
