UPSC NDA Math Model Paper 4

Given: Equation of sphere is x2+y2+z2+2x−2y−6z −14=0
By comparing the given equation with x2+y2+z2+2ux+2vy +2wz+d=0, we get
⇒u=1,v=−1,w=−3 and d=−14
As we know that, equation x2+y2+z2+2ux+2vy +2wz+d=0 represents a sphere with centre (−u,−v,−w) and radius r=√u2+v2+w2−d
So, the centre of given sphere is: (- 1, 1, 3)
By substituting u=1,v=−1,w=−3 and d=−14 in the equation r=√u2+v2+w2−d we get,
⇒r=√12+(−1)2+(−3)2−(−14) =5
Hence, the centre is (- 1, 1, 3) and radius is 5 units