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AP EAMCET Engineering 2011 Solved Paper

Section: Mathematics

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Question : 160 of 160

Marks: +1, -0

The radius of the circle given by

x^{2}+y^{2}+z^{2}+2x-2y

-4z-19=0

=x+2y+2z+7

, is

4
3
2
1

Solution:

Equation of sphere,

x^{2}+y^{2}+z^{2}+2x-2y-4z-19=0

.......(ii)

and equation of plane

x+2y+2z+7=0

.......(ii)

centre of the sphere

(-1,1,2)

and radius of the sphere

R=\sqrt{1+1+4+19}

=\sqrt{25}=5

Now,

p=

perpendicular distance from the centre to the plane

p=\frac{|-1+2+4+7|}{\sqrt{1+4+4}}=\frac{12}{\sqrt{9}}

=\frac{12}{3}=4

In

\Delta AOC

R^{2}=p^{2}+r^{2}

⇒

r^{2}=R^{2}-p^{2}

⇒

r^{2}=25-16=9

⇒

r=3

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