The equation of the sphere with centre (3, 6, -4) and touching the plane 2x -2y -z -10 = 0 is

x^{2} + y^{2} + z^{2} - 6 x + 12 y - 8 z + 45 = 0

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

x^{2} + y^{2} + z^{2} + 6 x - 9 y + 8 z + 35 = 0

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

x^{2} + y^{2} + z^{2} - 6 x + 9 y + 8 z + 35 = 0

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

x^{2} + y^{2} + z^{2} - 6 x - 12 y + 8 z + 45 = 0

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

Open in App

Solution

The correct option is C $x^{2} + y^{2} + z^{2} - 6 x - 12 y + 8 z + 45 = 0$
Radius of the sphere = Distance of (3, 6, -4) from the given plane
$= \frac{| 2 \times 3 - 2 \times 6 + 4 - 10}{{\sqrt{2^{2} + (- 2)^{2} +}}^{2}} = \frac{12}{3} = 4$
Hence, the required equation of the sphere is $(x - 3)^{2} + (y - 6)^{2} + (z + 4)^{2} = (4)^{2}$
or $x^{2} + y^{2} + z^{2} - 6 x - 12 y + 8 z + 45 = 0$

Suggest Corrections

Similar questions

The equation of circle concentric with circle $x^{2} + y^{2} - 6 x + 12 y + 15$ = 0 and double its area is

x^{2} + y^{2} + z^{2} - 4 x + 6 y - 8 z - 7 = 0 ?

x^{2} + y^{2} + z^{2} + 6 y + 2 z + 8 = 0

x^{2} + y^{2} + z^{2} + 6 x + 8 y + 4 z + 20 = 0

x^{2} + y^{2} + z^{2} - 2 x + 3 y - 8_{Z} - 1 = 0

(- 1, 2, 0)

x^{2} + y^{2} + z^{2} - 4 x + 6 y + 8 z + 4 = 0

^{'} z^{'}

Join BYJU'S Learning Program