Question 440 of 521MediumMCQ
The plane x+2y−z=4 cuts the sphere x2+y2+z2−x+z−2=0 in a circle of radius Solutions · 1
Official solution
Perpendicular distance of center \[\left( {{1 \over 2},0, - {1 \over 2}} \right)\]
from x+2y−2=4 is given by
\[{{\left| {{1 \over 2} + {1 \over 2} - 4} \right| = \sqrt {{3 \over 2}} } \over {\sqrt 6 }}\]
radius of sphere
\[ = \sqrt {{1 \over 4} + {1 \over 4} + 2} = \sqrt {{5 \over 2}} \]
\[\therefore\] radius of circle
\[ = \sqrt {{5 \over 2} - {3 \over 2}} = 1.\] Answer the question to unlock