    Question

# The equation of sphere passing through points (1,0,0),(0,1,0) and (0,0,1) and having minimum radius, is

A
3(x2+y2+z2)xyz1=0
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B
3(x2+y2+z2)2(x+y+z)1=0
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C
3(x2+y2+z2)+2(x+y+z)+1=0
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D
3(x2+y2+z2)+x+y+z+1=0
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Solution

## The correct option is B 3(x2+y2+z2)−2(x+y+z)−1=0Let the equation of sphere be x2+y2+z2+2ux+2vy+2wz+d=0⋯(i) As the sphere passes through points (1,0,0),(0,1,0) and (0,0,1), we get 1+2u+d=0,1+2v+d=0 and 1+2w+d=0 ⇒u=v=w=−12(d+1) If R is the radius of sphere, ⇒R2=u2+v2+w2−d =34(d+1)2−d =34[d2+2d+1−43d] =34[(d+13)2+89] So, for minimum radius, d+13=0 ⇒d=−13⇒u=v=w=−12(1−13)=−13 Putting these values in equation (i), we get x2+y2+z2−23(x+y+z)−13=0 ⇒3(x2+y2+z2)−2(x+y+z)−1=0  Suggest Corrections  6      Similar questions
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