Equation of tangent plane to sphere?

Question

Find equations of the tangent planes to the sphere x^2 - 6x + y^2 - 8y + z^2 + 2z = 10 with a normal vector <2,2,1>. Hint, first find the center and radius of the sphere and use the normal to go from the center to the two points of tangency that are on the sphere.

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So I found the center and radius (3,4,-1) r=6
The length of the normal is 3 so I need to multiply the vector by 2 to find the points.

But how would the equations look?
Is it just <±t+4x, ±t+4y, ±t-z> = ?
So that any number t would find the point?

Answer 1

Find equations of the tangent planes to the
sphere x² - 6x + y² - 8y + z² + 2z = 10 with normal vector
v = <2,2,1>.

Find the center and radius of the sphere by putting the equation in standard form.

x² - 6x + y² - 8y + z² + 2z = 10
(x² - 6x + 9) + (y² - 8y + 16) + (z² + 2z + 1) = 10 + 9 +16 + 1
(x - 3)² + (y - 4)² + (z + 1)² = 36

The center of the sphere is (h, k, p) = Q(3, 4, -1).
The radius is √36 = 6.

The magnitude of the directional vector of the line is:

|| v || = √(2² + 2² + 1²) = √(4 + 4 + 1) = √9 = 3

The desired line goes thru the center of the sphere with direction v = <2, 2, 1>. The equation of the line is:

L(t) = Q + tv = <3, 4, -1> + t<2, 2, 1>.

The line will intersect the sphere at a distance of 6 in either direction from the center of the sphere. That is a directed distance of ±2v.

The points of intersection A and B are:

A = Q + 2v = <3, 4, -1> + 2<2, 2, 1>
A = <3+2*2, 4+2*2, -1+2*1> = A(7, 8, 1)

B = Q - 2v = <3, 4, -1> - 2<2, 2, 1>
B = <3-2*2, 4-2*2, -1-2*1> = B(-1, 0, -3)

Now write the equations of the planes tangent to the sphere at A and B. The directional vector v, of the line is also the normal vector of the tangent planes. Use the normal vector and the points of tangency to write the equations of the tangent planes.

A(7, 8, 1)
v = <2, 2, 1>

The first plane is:

2(x - 7) + 2(y - 8) + 1(z - 1) = 0
2x - 14 + 2y - 16 + z - 1 = 0
2x + 2y + z - 31 = 0
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B(-1, 0, -3)
v = <2, 2, 1>

The second plane is:

2(x + 1) + 2(y - 0) + 1(z + 3) = 0
2x + 2 + 2y - 0 + z + 3 = 0
2x + 2y + z + 5 = 0
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