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]

• A. 2
• B.
  
• C. 3
• D. 1

Answer 1

The correct option is D

1

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, $r=\sqrt{R^2-P^2}=\sqrt{\frac{5}{2}-\frac{3}{2}}=1$.