Three Dimensional Geometry Question 39
Question: 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]
Options:
A) 2
B) $ \sqrt{2} $
C) 3
D) 1
Show Answer
Answer:
Correct Answer: D
Solution:
Perpendicular distance to centre $ ( \frac{1}{2},0,-\frac{1}{2} ) $ from $ x+2y-z=4 $ is, $ P=\frac{| \frac{1}{2}+\frac{1}{2}-4 |}{\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 $ .
BETA
