The equation of the sphere whose centre is (6,-1, text 2) and which touches the plane 2x-y+2z-2=0, is:

Question

The equation of the sphere whose centre is (6,-1, text 2) and which touches the plane 2x-y+2z-2=0, is: (A)  x2+y2+z2-12x+2y-4z-16=0  (B)  x2+y2+z2-12x+2y-4z=0  (C)  x2+y2+z2-12x+2y-4z+16=0  (D)  x2+y2+z2-12x+2y-4z+6=0

Tardigrade Admin · Accepted Answer

Radius of sphere is J- distance of (6,-1,2) from 2x-y+2z-2=0 i.e., | (12+1+4-2/√4+1+4) |=| (15/3) |=5 ∴ Equation of sphere is (x-6)2+(y+1)2+(z-2)2=25 ⇒ x2+y2+z2-12x+2y-4z +36+1+4-25=0 ⇒ x2+y2+z2-12x+2y-4z+16=0

Q. The equation of the sphere whose centre is $(6, - 1, 2)$ and which touches the plane $2 x - y + 2 z - 2 = 0,$ is:

$x^{2} + y^{2} + z^{2} - 12 x + 2 y - 4 z - 16 = 0$

$x^{2} + y^{2} + z^{2} - 12 x + 2 y - 4 z = 0$

$x^{2} + y^{2} + z^{2} - 12 x + 2 y - 4 z + 16 = 0$

$x^{2} + y^{2} + z^{2} - 12 x + 2 y - 4 z + 6 = 0$

$x^{2} + y^{2} + z^{2} - 12 x + 2 y - 4 z - 5 = 0$

Q. The equation of the sphere whose centre is (6,−1,2) and which touches the plane 2x−y+2z−2=0, is:

x2+y2+z2−12x+2y−4z−16=0

x2+y2+z2−12x+2y−4z=0

x2+y2+z2−12x+2y−4z+16=0

x2+y2+z2−12x+2y−4z+6=0

x2+y2+z2−12x+2y−4z−5=0