Find an equation of the plane tangent to the sphere at the pt where the ray pierces the sphere?

Question

A ray from the center of the sphere with equation x^+y^2+z^2+2x-8y-4z-15=0 passes through the pt P(7,3,-2). Find an equation of the plane tangent to the sphere at the pt where the ray pierces the sphere. Note that P is not a pt on the sphere or on the plane

Answer 1

The equation of the sphere is

x² + y² + z² + 2x - 8y - 4z - 15 = 0
x² + 2x + y² - 8y + z² - 4z = 15

Complete the squares of x, y, and z.

(x² + 2x +1) + (y² - 8y + 16) + (z² - 4z + 4) = 15 + 1 + 16 + 4
(x + 1)² + (y - 4)² + (z - 2)² = 36

This the equation of a sphere with center (-1, 4, 2) and
radius 6.

The ray goes thru two points O(-1, 4, 2) and P(7, 3, -2).

The directional vector v of the ray is:

v = OP = <7+1, 3-4, -2-2> = <8, -1, -4>

The magnitude of v is:

|| v || = √[8² + (-1)² + (-4)²] = √(64 + 1 + 16) = √81 = 9

We want to go a distance of 6 from O in the direction of v.

The point Q on the surface of the sphere is:

Q = O + (6/9)v = (-1, 4, 2) + (2/3)*<8, -1, -4>
Q = (-1 + 16/3, 4 - 2/3, 2 - 8/3) = (13/3, 10/3, -2/3)

The vector v is the normal vector of the plane tangent to the sphere at Q. The equation of the tangent plane is:

8(x - 13/3) - 1(y - 10/3) - 4(z + 2/3) = 0
8x - 104/3 - y + 10/3 - 4z - 8/3 = 0
8x - y - 4z - 102/3 = 0
8x - y - 4z - 34 = 0

Answer 2

[x+1]* + [y-4]* + [z-2]* = 15 + 1 + 16 + 4 =36.
Let the centre be A.
P - O = (8, -1, -4), with magnitude sqrt[81] = 9
If k[P-O] has magnitude 6, then clearly k = 2/3. [or -2/3, but I'll ignore that case for now].

Thus the tangent point is
O + (16/3, -2/3, -4/3) = (13, 10, 2)/3.
Call that point T.
The equation of your desired plane with general point X is
X . O = T. O
so
X. (-1, 4, 2) = (13, 10, 2) . (-1, 4, 2)/3
-x + 4y + 2z = [-13 + 40 + 4]/3 = 31/3.
For the other point of intersection, use -31/3. Anyway, I often add poorly so you should follow the method and check that this is coherent.