JEE Main & Advanced
Mathematics
Three Dimensional Geometry
Question Bank
Sphere
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, \[r=\sqrt{{{R}^{2}}-{{P}^{2}}}=\sqrt{\frac{5}{2}-\frac{3}{2}}=1\].