Question

How do i proove that a plane is a tangent plane of a sphere, when I am given equation of the plane and equation of the sphere?

Answer 1

gradiant of sphere gives the normal vector to plane

Answer 2

say, f = \(x^2 + y^2+z^2-r^2\) is a sphere,

then grad(f) = \(\left<2x, 2y, 2z\right>\) gives the normal vector to tangent at point \((x, y, z)\)

Answer 3

my circle is (x+1)^2 + (y-3)^2 + z^2 = 25 or the in other words circle centre is at (-1,3,0) and the radius is 5

Answer 4

if i differentiate the thing, I get 2x+2y+2z + 2 - 6

Answer 5

If my plane has a normal of (4,0,3) if I dot product it with (2,2,2)?? it wouldn't give a zero? I'm not quite sure of what you mean yet. My plane equation is 4x+3z+29=0

Answer 6

use below :

if the plane is tangent, then the distance from center to plane must equal the radius

Answer 7

Find : distance from the center of sphere (-1, 3, 0) to the plane 4x + 3z + 29

Answer 8

if that equals the radius of sphere, then that plane is a tangent plane

Answer 9

distance from the center of sphere (-1, 3, 0) to the plane 4x + 3z + 29 :

\(\large \frac{4(-1) + 29}{\sqrt{4^2 + 3^2} } = 5\)

so yes, the given plane is a tangent plane