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Question


Equation to the sphere which passes through the circle x2+y2+z2=5,x+2y+3z=3 and touch the plane 4x+3y− 15=0

A
x2+y2+z2+2x+4y+6z+11=0
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B
x2+y2+z22x4y+6z11=0
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C
x2+y2+z22x4y6z+11=0
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D
x2+y2+z2+2x+4y+6z11=0
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Solution

The correct option is B x2+y2+z2+2x+4y+6z11=0
Equation to the sphere passes through the circle x2+y2+z2=5,x+2y+3z=3 is given by,
x2+y2+z25+λ(x+2y+3z3)=0
x2+y2+z2+λx+2λy+3λz(3λ+5)=0
Given plane 4x+3y15=0 touches this sphere,
∣ ∣ ∣4(λ2)+3(λ)155∣ ∣ ∣=(λ2)2+λ2+(3λ2+3λ+5)2
5λ26λ8=0λ=2,4/5
Hence, the sphere is x2+y2+z2+2x+4y+6z11=0
There will be one more sphere corresponding to another value of λ.

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