JEE Main · 2005 · 08 May 2005 · Offline · Three Dimensional Geometry

The plane

x + 2 y - z = 4

cuts the sphere

x^{2} + y^{2} + z^{2} - x + z - 2 = 0

in a circle of radius

(A)3
(B)1✓ Correct
(C)2
(D) $2$

Answer: (B)

Solution

Completing the square, the sphere

x^{2} + y^{2} + z^{2} - x + z - 2 = 0

has centre

(\frac{1}{2}, 0, - \frac{1}{2})

and radius

\frac{5}{2}

. The perpendicular distance from this centre to the plane

x + 2 y - z = 4

\frac{∣ \frac{1}{2} + 0 + \frac{1}{2} - 4∣}{1 + 4 + 1} = \frac{3}{6}

, so

d^{2} = \frac{3}{2}

. The radius of the circle of intersection is

R^{2} - d^{2} = \frac{5}{2} - \frac{3}{2} = 1

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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