Chapter 11 - Vectors and Vector-Valued Functions - 11.2 Vectors in Three Dimensions - 11.2 Exercises - Page 778: 25

$(x+2)^2 + y^2 + (z-4)^2 \leq 1^2$

Equation of sphere with radius $r$ and centered at $(a,b,c)$: $(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2$ Plug in center and radius: $(x+2)^2 + (y-0)^2 + (z-4)^2 = 1^2$ Because it's a ball, the equation must account for the fact that the sphere is not hollow. $(x+2)^2 + (y-0)^2 + (z-4)^2 \leq 1^2$