question_answer

The plane \[x+2y-z=4\] cuts the sphere \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}\] \[-x+z-2=0\] in a circle of radius [AIEEE 2005]

A)            2

B)            \[\sqrt{2}\]

C)            3

D)            1

Correct Answer: D

Solution :

                   Perpendicular distance to centre \[\left( \frac{1}{2},0,-\frac{1}{2} \right)\] from \[x+2y-z=4\]is,                    \[P=\frac{\left| \frac{1}{2}+\frac{1}{2}-4 \right|}{\sqrt{6}}=\sqrt{\frac{3}{2}}\] and radius of sphere \[R=\sqrt{\frac{5}{2}}\],                    So, \[r=\sqrt{{{R}^{2}}-{{P}^{2}}}=\sqrt{\frac{5}{2}-\frac{3}{2}}=1\].