How could I find the points on surface x^2 + y^2 + z^2 - 4x - 2y + 2z =3 where the tangent plane is horizontal

Question

possible points on the surface.....

a. (2, 1, 3), (2, 1, -5)

b. (-2, 1, 2), (-2, 1, -4)

c. (2, 1, 2), (2, 1, -4)

d. (-2, -1, 5), (-2, -1, -3)

e. (2, -1, 3), (2, -1, -5)

f. (2, -1, 2), (2, -1, -4)

or could it be none of these?

Answer 1

If horizontal means that the plane is orthogonal to the z axis, then the answer is a).

Differentiate partially wrt to x and y and set both equal to zero.

wrt x:
2x - 4 = 0
x = 2

wrt y:
2y - 2 = 0
y = 1

Sub x = 2 and y = 1 into original equation to find the two values of z. Those will be the two points where fx and fy = 0 ie tangent planes are orthogonal to z axis.

Answer 2

Complete the squares to writh the equation of the ellipsoid.

x² + y² + z² - 4x - 2y + 2z = 3
(x² - 4x) + (y² - 2y) + (z² + 2z) = 3
(x² - 4x + 4) + (y² - 2y + 1) + (z² + 2z + 1) = 3 + 4 + 1 + 1
(x - 2)² + (y - 1)² + (z + 1)² = 9

The center of the sphere is (2, 1, -1) and the radius is 3.

If by horizontal you mean perpendicular to the z axis the answer is:

(2, 1, -1 + 3) and (2, 1, -1 - 3)
(2, 1, 2) and (2, 1, -4)

The answer is c.