A conicoid given by equation ax2+by2+cz2+2fyz+2gxz+2hxy+2ux+2vy+2wz+d=0 has a point (x0,y0,z0) as a center if
For 1) a=b=c=1, u=v=w=1/2, d=-1, f=g=h=0:
x2+y2+z2+x+y+z=1 - the central conicoid have a center (-1/2,-1/2,-1/2)
For 2) a=2, h=2, g=1/2, u=-1/2, v=-3/2, w=5/2, d=3, b=c=f=0:
2x2+4xy+xz-x-3y+5z+3=0 - the conicoid not central
For 3) a=1, b=c=-1, h=1/2, f=2, u=1/2, d=g=v=w=0:
"\\begin{cases}\nx_0=-1\/2-z_0\/4 \\\\\n-1\/4-z_0\/8-z_0\/2+2z_0=0 \\\\\ny_0=z_0\/2\n\\end{cases}"
"\\begin{cases}\nx_0=-6\/11 \\\\\nz_0=2\/11 \\\\\ny_0=1\/11\n\\end{cases}"
x2-y2-z2+xy+4yz+x=0 - the central conicoid have a center (-6/11,1/11,2/11)
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