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Question

Obtain the equation of the sphere having the circle x2+y2+z2+10y4z=8, x+y+z=3 as a great circle.

A
x2+y2+z24x+6y8z+4=0
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B
x2+y2+z24x+6y8z+9=0
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C
x2+y2+z24x6y8z+4=0
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D
x2+y2+z24x6y8z+9=0
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Solution

The correct option is A x2+y2+z24x+6y8z+4=0
Any sphere through the intersection of the given circle
x2+y2+z2+10y4z8=0 ...(1)
x+y+z=3 ...(2)
is x2+y2+z2+10y4z8+k(x+y+z3)=0
x2+y2+z2+kx+(10+k)y(4k)z83k=0 ...(3)
Now if the circle of intersection of (1) and (2) is a greater circle of sphere (3), then
C(k2,(10+k)2,4k2) of sphere (3) must lie on the plane (2) i.e., we must have
k2(10+k2)+4k23=0
k10k+4k6=03k12=0
k=4
Substitute k=4 in (3), the required sphere is x2+y2+z24x+6y8z+4=0.

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