## Thomas' Calculus 13th Edition

circle with equation $x^2+z^2=9$ in the plane $y=-4$
Since, we have two equations shows the intersection between the equation of sphere $x^2+y^2+z^2=25$ and plane $y=-4$. It can be rewritten as: $x^2+(-4)^2+z^2=25$ or, $x^2+z^2=25-16$ and $x^2+z^2=9$ Thus, we have the set of points a circle with equation $x^2+z^2=9$ in the plane $y=-4$ , center: $(0,-4,0)$ with circle of radius $3$ .