Three Dimensional Geometry Question 73
Question: The radius of the circle in which the sphere $ x^{2}+y^{2}+z^{2}+2z-2y-4z-19=0 $ is cut by the plane $ x+2y+2z+7=0 $ is
Options:
A) 2
B) 3
C) 4
D) 1
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Center of the sphere is $ ( -1,1,2 ) $ and its radius $ =\sqrt{1+1+4+19}=5 $ CL, perpendicular distance of C from plane, is $ | \frac{-1+2+4+7}{\sqrt{1+4+4}} |=4 $ Now $ AL^{2}=CA^{2}-CL^{2}=25-16=9 $ Hence, radius of the circle $ =\sqrt{9}=3 $
BETA
