Three Dimensional Geometry Question 40
Question: The radius of sphere $ x+2y+2z=15 $ and $ x^{2}+y^{2}+z^{2}-2y-4z=11 $ is
[AMU 2005]
Options:
A) 2
B) $ \sqrt{7} $
C) 3
D) $ \sqrt{5} $
Show Answer
Answer:
Correct Answer: B
Solution:
Equation of sphere is, $ x^{2}+y^{2}+z^{2}-2y-4z=11 $ Centre of sphere = (0, 1, 2) and radius of sphere = 4 Let centre of circle be $ (\alpha ,\beta ,\gamma ) $ The d.r?s of line joining from centre of sphere to the centre of circle is $ (\alpha -0,\beta -1,\gamma -2) $ or $ (\alpha ,\beta -1,\gamma -2) $ But this line is normal at plane $ x+2y+2z=15 $ \ $ \frac{\alpha }{1}=\frac{\beta -1}{2}=\frac{\gamma -2}{2}=k $ $ \alpha =k,\beta =2k+1,\gamma =2k+2 $ $ \because $ Centre of circle lies on $ x+2y+2z=15 $ \ $ k+2(2k+1)+2(2k+2)=15 $
$ \Rightarrow k=1 $ So, centre of circle = (1, 3, 4) Therefore, Radius of circle $ =\sqrt{{{(Radius,of,sphere)}^{2}}-{{(Length,of,joining,line,of,centre)}^{2}}} $ $ =\sqrt{{{(4)}^{2}}-[{{(1-0)}^{2}}+{{(3-1)}^{2}}+{{(4-2)}^{2}}}] $ $ =\sqrt{16-9}=\sqrt{7} $ .
BETA
