Chapter 12 - Vectors and the Geometry of Space - 12.1 Three-Dimensional Coordinate Systems - 12.1 Exercises - Page 837: 24

$(x-5)^2+(y-4)^2+(z-9)^2=16$

The equation is of the form $(x-5)^2+(y-4)^2+(z-9)^2=r^2$ due to the center being $(5,4,9)$. The sphere will be contained in the first octant is all three variables are $\geq 0 $. If $y \leq 0$, the expression must be at least 16, so if we set $r^2=16$, the sphere is as large as possible and is completely contained in the first octant.