Find the centre of the sphere
$x^{2} + y^{2} + z^{2} + 2 z - x = 0$

(1, 0, - 1)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(0, 0, 0)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(1, 0, 1)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(- 1, 0, 1)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $(1, 0, - 1)$
Given equation of the sphere is
$x^{2} + y^{2} + z^{2} + 2 (z - x) = 0$
$x^{2} - 2 x + 1 + y^{2} + z^{2} + 2 z + 1 = 1 + 1 {(x - 1)}^{2} + {(y - 0)}^{2} + {(z + 1)}^{2} = {(\sqrt{2})}^{2}$
From this equation, we can say that the coordinates of centre of the sphere is $(1, 0, - 1) .$

Suggest Corrections

Similar questions

Find the centre of the sphere
$x^{2} + y^{2} + z^{2} + 2 z - x = 0$

x^{2} + y^{2} + z^{2} + 2 x - 4 y - 6 z + 5 = 0

2 x^{2} + 2 y^{2} + 2 z^{2} - 2 x + 4 y + 2 z + 3 = 0

x^{2} + y^{2} + z^{2} - 3 x + 4 y - 2 z - 5 = 0

5 x - 2 y + 4 z + 7 = 0

(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) +

(z - z_{1}) (z - z_{2}) = 0

If the radius of the sphere
$x^{2} + y^{2} + z^{2} - 2 x - 4 y - 6 z$ = 0
is r, then find the value of $r^{2}$

Join BYJU'S Learning Program