Question

The radius of the circle in which the sphere

$x ^ { 2 } + y ^ { 2 } + z ^ { 2 } + 2 x - 2 y - 4 z - 19 = 0$is cut by the plane

$x + 2 y + 2 z + 7 = 0$is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

3

Answer 2

For sphere $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } + 2 x - 2 y - 4 z - 19 = 0$ ,

Centre O is (–1, 1, 2) and radius $= \sqrt { 1 + 1 + 4 + 19 } = 5$,

Now, OL = length of perpendicular from O to plane

$x + 2 y + 2 z + 7 = 0$ is

$= \frac { - 1 + 2 + 4 + 7 } { \sqrt { 1 + 4 + 4 } } = \frac { 12 } { 3 } = 4$ , i.e. $O L = 4$ .

In $\triangle O L B , L B = \sqrt { O B ^ { 2 } - O L ^ { 2 } } = \sqrt { 25 - 16 } = 3$