The radius of the sphere x2+y2+z2=12 x+4 y+3 z is

Question

The radius of the sphere x2+y2+z2=12 x+4 y+3 z is (A) (13/2) (B) 13 (C) 26 (D) 52

Tardigrade Admin · Accepted Answer

Given equation of sphere is x2+y2+z2-12 x-4 y-3 z=0 ∴ Centre of sphere is (6,2, (3/2)). ∴ Radius of sphere =√(6)2+(2)2+((3/2))2 =√36+4+(9/4) =√(169/4)=(13/2)

Q. The radius of the sphere $x^{2} + y^{2} + z^{2} = 12 x + 4 y + 3 z$ is

$\frac{13}{2}$

13

26

52

Given equation of sphere is
$x^{2} + y^{2} + z^{2} - 12 x - 4 y - 3 z = 0$
$∴$ Centre of sphere is $(6, 2, \frac{3}{2})$ .
$∴$ Radius of sphere $= (6)^{2} + (2)^{2} + (\frac{3}{2})^{2}$
$= 36 + 4 + \frac{9}{4}$
$= \frac{169}{4} = \frac{13}{2}$

Q. The radius of the sphere x2+y2+z2=12x+4y+3z is

213​

13

26

52

Given equation of sphere is x2+y2+z2−12x−4y−3z=0 ∴ Centre of sphere is (6,2,23​). ∴ Radius of sphere =(6)2+(2)2+(23​)2​ =36+4+49​​ =4169​​=213​

Q. The radius of the sphere $x^{2} + y^{2} + z^{2} = 12 x + 4 y + 3 z$ is

$\frac{13}{2}$

Given equation of sphere is
$x^{2} + y^{2} + z^{2} - 12 x - 4 y - 3 z = 0$
$∴$ Centre of sphere is $(6, 2, \frac{3}{2})$ .
$∴$ Radius of sphere $= (6)^{2} + (2)^{2} + (\frac{3}{2})^{2}$
$= 36 + 4 + \frac{9}{4}$
$= \frac{169}{4} = \frac{13}{2}$