The equation of the sphere having center at $(3, 4, 5)$ and touching the $x y$ -plane

x^{2} + y^{2} + z^{2} - 6 x - 8 y - 10 z + 25 = 0

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

x^{2} + y^{2} + z^{2} - 6 x - 8 y - 10 z + 34 = 0

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

x^{2} + y^{2} + z^{2} - 6 x - 8 y - 10 z + 41 = 0

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

x^{2} + y^{2} + z^{2} - 6 x - 8 y - 10 z + 50 = 0

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

Open in App

Solution

The correct option is A $x^{2} + y^{2} + z^{2} - 6 x - 8 y - 10 z + 25 = 0$
Given centre of the sphere is $(3, 4, 5)$ .
Since the sphere is touching the $x y$ -plane, therefore radius of the sphere is same as the magnitude of $z$ -coordinate of the centre of the sphere which is $5$ .
Hence, equation of the required sphere is
$(x - 3)^{2} + (y - 4)^{2} + (z - 5)^{2} = 5^{2}$
$\Rightarrow x^{2} + y^{2} + z^{2} - 6 x - 8 y - 10 z + 25 = 0$

Suggest Corrections

Similar questions

x^{2} + y^{2} + z^{2} - 6 x + + 8 y - 10 z + 1 = 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

2 a x - 3 a y + 4 a z + 6 = 0

x^{2} + y^{2} + z^{2} + 6 x - 8 y - 2 z = 13

x^{2} + y^{2} + z^{2} - 10 x + 4 y - 2 z = 8

a =

2 a x - 3 a y + 4 a z + 6 = 0

x^{2} + y^{2} + z^{2} + 6 x - 8 y - 2 z = 13

x^{2} + y^{2} + z^{2} - 10 x + 4 y - 2 z = 8

a

Join BYJU'S Learning Program