The centre of the sphere $x^{2} + y^{2} + z^{2} - a x - b y - c z = 0$ is.

(a, b, c)

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

(- a, - b, - c)

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

(\frac{a}{2}, \frac{b}{2}, \frac{c}{2})

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

(- \frac{a}{2}, - \frac{b}{2}, - \frac{c}{2})

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

Open in App

Solution

The correct option is B $(\frac{a}{2}, \frac{b}{2}, \frac{c}{2})$
A sphere with center $(l, m, n)$ and radius $r$ is given by the equation $(x - l)^{2} + (y - m)^{2} + (z - n)^{2} = r^{2}$
When expanded, $x^{2} - 2 l x + l^{2} + y^{2} - 2 m y + m^{2} + z^{2} - 2 n z + n^{2} = r^{2}$ , this can also be written as $x^{2} + y^{2} + z^{2} - 2 l x - 2 m y - 2 n z + l^{2} + m^{2} + n^{2} - r^{2} = 0$
If compared to the given equation $x^{2} + y^{2} + z^{2} - a x - b y - c z = 0$ , we have the center of the sphere to be
$(\frac{a}{2}, \frac{b}{2} . \frac{c}{2})$

Suggest Corrections

Similar questions

a^{2} + b^{2} + c^{2} = 2, x^{2} + y^{2} + z^{2} = 2

a, b, c, x, y, z > 0

(a + b + c)^{3} - 4 (b + c) (c + a) (a + b) + 5 a b c = 0

a x^{2} + b y^{2} + c z^{2} = a x + b y + c z = y z + z x + x y = 0

a x + b y + c z = a^{2} x + b^{2} y + c^{z} = 0, x + y + z + (b - c) (c - a) (a - b) = 0 .

If the centre of the sphere $x^{2} + y^{2} + z^{2} - 2 x - 4 y - 6 z = 0$ is (a,b,c) , find the value of a+b+c

a x^{2} + b y^{2} + c z^{2} + 2 f y z + 2 g x z + 2 h x y + 2 u x + 2 v y + 2 w z + d = 0

Join BYJU'S Learning Program