The plane x + 2y - z = 4 cuts the sphere x2 + y2 + z2-x + z- 2= 0 in a circle of radius:

Question

The plane x + 2y - z = 4 cuts the sphere x2 + y2 + z2-x + z- 2= 0 in a circle of radius: (A) √2 (B) 2 (C) 1 (D) 3

Tardigrade Admin · Accepted Answer

Since the centre of sphere x2+y2+z2-x+z-2=0 is ((1/2), 0, -(1/2)) and radius of sphere =√(1/4)+(1/4)+2=(√10/2) Distance of plane from centre of sphere =|((1/2)+(1/2)-4/√1+4+1)|=(3/√6) So radius of circle =√(10/4)-(9/6) =√(30-18/12)=√(12/12)=1

Q. 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:

$2$

$2$

$1$

$3$

Q. The plane x+2y−z=4 cuts the sphere x2+y2+z2−x+z−2=0 in a circle of radius:

2​

2

1

3

Since the centre of sphere x2+y2+z2−x+z−2=0 is (21​,0,−21​) and radius of sphere =41​+41​+2​=210​​ Distance of plane from centre of sphere =∣∣​1+4+1​21​+21​−4​∣∣​=6​3​ So radius of circle =410​−69​​ =1230−18​​=1212​​=1

Q. 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:

$2$

$2$

$1$

$3$