Chapter 12 - Section 12.6 - Cylinders and Quadric Surfaces - 12.6 Exercises - Page 840: 38

$$(4x-12)^2 + (y-4)^2 = (z+2)^2 - 211$$

$4x^2 + y^2 + z^2 - 24x - 8y + 4z = -55$ $4x^2 - 24x + y^2 - 8y + z^2 + 4z = -55$ $(4x^2 - 24x + 144) + (y^2 - 8y + 16) + (z^2 + 4z - 4) = -55 - 144 - 16 + 4$ $(4x^2 - 24x + 144) + (y^2 - 8y + 16) + (z^2 + 4z - 4) = -211$ $(4x - 12)^2 + (y - 4)^2 + (z + 2)^2 = -211$ $(4x - 12)^2 + (y - 4)^2 = (z + 2)^2 - 211$