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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 ?

Arthur
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  • A quick use of geogebra gives $(-\frac15,\frac12,-\frac3{10}).$ Also in M2 R=QQ[x,y,z] s1=x^2+y^2+z^2+2*x+2*y+2*z+2 s2=x^2+y^2+z^2+4*y s3=x^2+y^2+z^2+3*x-2*y+8*z+6 s4=x^2+y^2+z^2-x+4*y-6*z-2 I=ideal(s1-s2,s1-s3,s1-s4) primaryDecomposition I -- {ideal (10z + 3, 2y - 1, 5x + 1)} But I suppose you'd like a proof that these differences all give the same ideal. – Jan-Magnus Økland Jan 30 '23 at 14:04
  • @Jan-MagnusØkland Exactly! But how to prove the required coordinates for the radical centre ? – Arthur Jan 30 '23 at 14:12

1 Answers1

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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".

Apass.Jack
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  • Actually this was the thing I was looking for. If we try to solve each of the equation by taking each of the radical axis, then truly it becomes huge. Your method correctly demonstrates a more faster way that going all of the equations two at a time. Thank you! – Arthur Jan 30 '23 at 16:30
  • You are welcome! – Apass.Jack Jan 30 '23 at 16:31
  • @Franklin R=QQ[k][x,y,z] I=ideal(s1-k,s2-k,s3-k,s4-k) primaryDecomposition I -- {ideal (50k - 119, 10z + 3, 2y - 1, 5x + 1)} So the radical centre is the only point at which the value is the same of the four normalised defining quadratics. – Jan-Magnus Økland Jan 30 '23 at 22:38