2∗u1∗u2+2∗v1∗v2+2∗w1∗w2=d1+d2 --- it is a orthogonality condition
x2+y2+z2+6∗y+2∗z+8=0 --- the first sphere
u1=0/2=0 ; v1=6/2=3 ; w1=2/2=1 ; d1=8
x2+y2+z2+6∗x+8∗y+4∗z+20=0 --- the second sphere
u2=6/2=3 ; v2=8/2=4 ; w2=4/2=2 ; d2=20
As the first sphere is orthogonal to the second sphere, we must have
2∗0∗3+2∗3∗4+2∗2∗1=8+20
24+4=28
28=28 - equality is true
so these two spheres are orthogonal
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