Find the equation of the two tangent planes to the sphere $x^{2} + y^{2} + z^{2} - 2 y - 6 z + 5 = 0$ which are parallel to the plane $2 x + 2 y - z = 0.$

2 x + 2 y - z + (2 \pm 3 \sqrt{3}) = 0.

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

2 x + 2 y - z + (2 \pm 3 \sqrt{5}) = 0.

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

2 x + 2 y - z + (1 \pm 3 \sqrt{5}) = 0.

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

2 x + 2 y - z + (1 \pm 3 \sqrt{3}) = 0.

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

Open in App

Solution

The correct option is C $2 x + 2 y - z + (1 \pm 3 \sqrt{5}) = 0.$
Any plane parallel to the given plane is
$2 x + 2 y - z + λ = 0$ ...(1)
If it is a tangent plane then perpendicular from centre $(0, 1, 3)$ should be equal to radius $\sqrt{(0 + 1 + 9 - 5)} = \sqrt{5}$ .
$∴ \frac{0 + 2.1 - 3 + λ}{\sqrt{4 + 4 + 1}} = \pm \sqrt{5}$
$∴ λ = 1 \pm 3 \sqrt{5}$
Hence, the required tangent planes are
$2 x + 2 y - z + (1 \pm 3 \sqrt{5}) = 0.$

Suggest Corrections

Similar questions

2 x + 2 y + z + 3 = 0

2 x + 2 y + z - 15 = 0

2 x - 2 y + z + 12 = 0

x^{2} + y^{2} + z^{2} - 2 x - 4 y + 2 z - 3 = 0

2 x - 2 y + z + 12 = 0

x^{2} + y^{2} + z^{2} - 2 x - 4 y + 2 z - 3 = 0

2 x + 2 y - z = k

x^{2} + y^{2} + z^{2} - 4 x + 2 y - 6_{Z} + 5 = 0

z

k =

Join BYJU'S Learning Program