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

• A. 2
• B. $\sqrt{2}$
• C. 3
• D. 1

Answer 1

Answer: (D)

1

Answer 2

Perpendicular distance to centre $\left( \frac { 1 } { 2 } , 0 , - \frac { 1 } { 2 } \right)$ from $x + 2 y - z = 4$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, $r = \sqrt { R ^ { 2 } - P ^ { 2 } } = \sqrt { \frac { 5 } { 2 } - \frac { 3 } { 2 } } = 1$.