The locus of the equation $x^{2} + y^{2} + z^{2} + 1 = 0$ is

An empty set

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

A sphere

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

A degenerate set

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

A pair of planes

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

Open in App

Solution

The correct option is A
An empty set

An empty set ( $∵ x^{2} + y^{2} + z^{2} + 1 = 0)$ is not possible for any real value of x)

Suggest Corrections

Similar questions

2 (x^{2} + y^{2} + z^{2}) + 2 x - 6 y + 4 z + 56 = 0

The equation of locus of a point where distance from the y-axis is equal to its distance from the point A(2,1,-1) is-

x^{2} + y^{2} + z^{2} = 0

x + y + z = 0

\frac{1}{x^{2} + y^{2} - z^{2}} + \frac{1}{y^{2} + z^{2} - x^{2}} + \frac{1}{z^{2} + x^{2} - y^{2}}

Join BYJU'S Learning Program