Answer
$x^{2}+y^{2}=16,\qquad z=3$
Work Step by Step
Plane:
perpendicular to the z-axis $\Rightarrow$ parallel to the xy-plane ($z=0)$
Passes through a point in which $z=3 \Rightarrow$ the plane is $z=3$ .
Sphere:$\qquad x^{2}+y^{2}+z^{2}=25$
Substitute z=3 into the equation for the sphere,
$x^{2}+y^{2}+9=25,\qquad z=3$
$x^{2}+y^{2}=16,\qquad z=3$
(circle of radius 4 in the plane $z=3$, centered on the z-axis).
