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Question: The radius of sphere \(x + 2y + 2z = 15\)and \(x^{2} + y^{2} + z^{2} - 2y - 4z = 11\) is...

The radius of sphere x+2y+2z=15x + 2y + 2z = 15and

x2+y2+z2−2y−4z=11x^{2} + y^{2} + z^{2} - 2y - 4z = 11 is

A

2

B

7\sqrt{7}

C

3

D

5\sqrt{5}

Answer

7\sqrt{7}

Explanation

Solution

Equation of sphere is, x2+y2+z2−2y−4z=11x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - 2 y - 4 z = 11

Centre of sphere = (0, 1, 2) and radius of sphere = 4

Let centre of circle be(α,β,γ)( \alpha , \beta , \gamma )

The d.r’s of line joining from centre of sphere to the centre of circle is (α−0,β−1,γ−2)( \alpha - 0 , \beta - 1 , \gamma - 2 )or (α,β−1,γ−2)( \alpha , \beta - 1 , \gamma - 2 )

But this line is normal at plane x+2y+2z=15x + 2 y + 2 z = 15

∴ α1=β−12=γ−22=k\frac { \alpha } { 1 } = \frac { \beta - 1 } { 2 } = \frac { \gamma - 2 } { 2 } = k

α=k,β=2k+1,γ=2k+2\alpha = k , \beta = 2 k + 1 , \gamma = 2 k + 2

∙∙\bullet \bullet Centre of circle lies on x+2y+2z=15x + 2 y + 2 z = 15

∴ k+2(2k+1)+2(2k+2)=15k + 2 ( 2 k + 1 ) + 2 ( 2 k + 2 ) = 15 ⇒k=1\Rightarrow k = 1

So, centre of circle = (1, 3, 4)

Therefore, Radius of circle

=( Radius of sphere )2−( Length of joining line of centre )2= \sqrt { ( \text { Radius of sphere } ) ^ { 2 } - ( \text { Length of joining line of centre } ) ^ { 2 } }

=(4)2−[(1−0)2+(3−1)2+(4−2)2]= \sqrt { ( 4 ) ^ { 2 } - \left[ ( 1 - 0 ) ^ { 2 } + ( 3 - 1 ) ^ { 2 } + ( 4 - 2 ) ^ { 2 } \right] } =16−9=7= \sqrt { 16 - 9 } = \sqrt { 7 } .