Answer
$x^{2}+y^{2}=16,\qquad z=3$
Work Step by Step
The plane:
- is perpendicular to the z-axis $\Rightarrow$ parallel to the xy-plane ($z=0)$
- passes through in which $z=3 \Rightarrow$ the plane is $z=3$ .
Substitute z=3 into the equation for the sphere,$\qquad x^{2}+y^{2}+z^{2}=25$
$x^{2}+y^{2}+9=25,\qquad z=3$
$x^{2}+y^{2}=16,\qquad z=3$
Thus, we get a circle in the plane $z=3$.
