Solveeit Logo
Login

Question

Question: The equation of the sphere touching the three co-ordinate planes is...

The equation of the sphere touching the three co-ordinate planes is

A

x2+y2+z2+2a(x+y+z)+2a2=0x^{2} + y^{2} + z^{2} + 2a(x + y + z) + 2a^{2} = 0

B

x2+y2+z2−2a(x+y+z)+2a2=0x^{2} + y^{2} + z^{2} - 2a(x + y + z) + 2a^{2} = 0

C

x2+y2+z2±2a(x+y+z)+2a2=0x^{2} + y^{2} + z^{2} \pm 2a(x + y + z) + 2a^{2} = 0

D

None of these

Answer

x2+y2+z2−2a(x+y+z)+2a2=0x^{2} + y^{2} + z^{2} - 2a(x + y + z) + 2a^{2} = 0

Explanation

Solution

Given, sphere touching the three co-ordinates planes. So clearly the centre is (a,a,a)( a , a , a ) and radius is a.

From (x−a)2+(y−b)2+(z−c)2=r2( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } + ( z - c ) ^ { 2 } = r ^ { 2 },

∴\therefore (x−a)2+(y−a)2+(z−a)2=a2( x - a ) ^ { 2 } + ( y - a ) ^ { 2 } + ( z - a ) ^ { 2 } = a ^ { 2 }

x2+y2+z2−2ax−2ay−2az+3a2=a2x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - 2 a x - 2 a y - 2 a z + 3 a ^ { 2 } = a ^ { 2 }

∴\therefore x2+y2+z2−2a(x+y+z)+2a2=0x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - 2 a ( x + y + z ) + 2 a ^ { 2 } = 0 is the required equation of sphere.