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UPSC NDA I 2014 Mathematics Paper

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For the next two (02) items that follow:

Consider the spheres

x^2+y^2+z^2-4y+3=0

and

x^2+y^2+z^2+2x+4z-4=0

[NDA I 2014]

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Question : 64 of 120

Marks: +1, -0

What is the distance between the centres of the two spheres ?

5 unit
4 unit
3 unit
2 unit

Solution:

x^2+y^2+z^2-4y+3=0

\Rightarrow x^2+(y-2)^2+z^2=1

Center≡ ( 0, 2, 0),Radius = 1

x^2+y^2+z^2+2x+4z-4=0

\Rightarrow (x+1)^2+y^2+(z+2)^2=9

Center ≡ (-1,0,-2) Radius = 3

Distance

=\sqrt{(-1,-0)^2+(0-2)^2(-2-0)^2}=\sqrt{1+4+4}=3

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