The centre and radius of the sphere $2 x^{2} + 2 y^{2} + 2 z^{2} - 2 x + 4 y + 2 z + 1 = 0$

(1 / 2, - 1, - 1 / 2), 1

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

(- 1 / 2, - 1, - 1 / 2), 2

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

(1 / 2, - 1, - 1 / 2), 2

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

(- 1 / 2, - 1, - 1 / 2), 1

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

Open in App

Solution

The correct option is A $(1 / 2, - 1, - 1 / 2), 1$
Given sphere is $2 x^{2} + 2 y^{2} + 2 z^{2} - 2 x + 4 y + 2 z + 1 = 0$
It can be written as,
$x^{2} + y^{2} + z^{2} - x + 2 y + z + 1 / 2 = 0$
$x^{2} + y^{2} + z^{2} - \frac{2 x}{2} + \frac{1}{4} - \frac{1}{4} + 2 y + 1 - 1 + \frac{2 z}{2} + \frac{1}{4} - \frac{1}{4} + 1 / 2 = 0$
${(x - \frac{1}{2})}^{2} + (y + 1)^{2} + {(z + \frac{1}{2})}^{2} - \frac{1}{2} - 1 + \frac{1}{2} = 0$
$⟹ (x - 1 / 2)^{2} + (y + 1)^{2} + (z + 1 / 2)^{2} = (1)^{2}$
Hence centre and radius of sphere are $(\frac{1}{2}, - 1, \frac{- 1}{2})$ and $1$ respectively.

Suggest Corrections

Similar questions

(1, - 1, 1)

2 x^{2} + 2 y^{2} + 2 z^{2} - 2 x + 4 y + 6 z - 1 = 0

(2, - \frac{1}{2}), (\frac{1}{2}, - \frac{1}{2}), (2, \frac{\sqrt{3} - 1}{2})

Which term of the G.P. :

(i) $\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2 \sqrt{2}}, \frac{1}{4 \sqrt{2}}, . . . . i s \frac{1}{512 \sqrt{2}}$ ?

(ii) 2, $2 \sqrt{2}, 4, . . . . i s 128$ ?

(iii) $\sqrt{3}, 3, 3 \sqrt{3}, . . . . . i s 729$ ?

(iv) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27} . . . i s \frac{1}{19683}$ ?

f (x) = \frac{x}{1 + x^{2}}

f

Join BYJU'S Learning Program