Question

Show that the plane `2x-4y-z+9=0` touches the sphere which passes through `(1,1,6)` and whose centre is `(2,-3,4)`. Also, find the point of contact.?

Answer 1

See below.

Explanation:

With `p = (x,y,z)`

`Pi -> << vec n, p - p_0 >> = 2x-4y-z+9=0`

here

`vec n = (2,-4,-1)`
`p_0 = (0,0,9)`

`S-> norm(p-p_1) = norm(p_1-p_2) = r`

here

`p_1=(2,-3,4)`
`p_2=(1,1,6)`

Now if `Pi` and `S` are tangent then

`p_t = p_1-r vec n/norm(vec n) in Pi`

Computing `p_t = (0,1,5)`

and as can easily be verified

`<< vec n , p_t-p_0 >> = 0` so the tangency point is

`p_t = (0,1,5)`

Answer 2

Distance from the Plane to Centre = Radius of the Given Sphere
Point of contact `(0, 1, 5)`

Explanation:

The Sphere passes through the point `(1, 1, 6)`.
Hence, The distance from it's centre `(2, -3, 4)` to this point is the radius of the sphere.

`:. r = sqrt((2-1)^2 + (-3-1)^2 + (4-6)^2)`
`=sqrt(21)`

Now, the distance between the given plane `2x-4y-z+9=0` to the centre of the sphere is :

`= |(2.2 + (-4).(-3) + (-1).4 + 9)|/(sqrt(2^2 + (-4)^2 + (-1)^2))`
`= |4+12-4+9|/(sqrt(4+16+1))`
`=21/sqrt21`
`=sqrt21`

As the distance from the plane to centre is equal to the distance of the radius of the sphere, it touches the sphere.

The point of contact is:
`vec u = (2, -3, 4)-(2, -4, -1)`
`= (0, 1, 5)`