Find the conical surface with vertex O through the curve…

Find the conical surface with vertex $O$ through the curve $x^2+y^2+(z-5)^2=9,\ z=4$.

Explanation of Method

The cone with vertex O through the circle cut from a sphere by a horizontal plane.

Setting $z=4$: $x^2+y^2=8$. So the directrix is the circle of radius $2\sqrt2$ at height $4$.

$OM$ meets $z=4$ at $\left(\tfrac{4x}z,\ \tfrac{4y}z,\ 4\right)$, so $\tfrac{16(x^2+y^2)}{z^2}=8$.

\[2(x^2+y^2)=z^2\]

This is a circular cone with axis $Oz$ and $\tan\alpha=\tfrac{2\sqrt2}{4}=\tfrac1{\sqrt2}$. Check: $(2,2,4)$ gives $16=16$. $\checkmark$

$O$ $(0,0,5)$

The sphere meets $z=4$ in the circle $x^2+y^2=8$ (red); the cone from $O$ through it is $2(x^2+y^2)=z^2$.

\[\boxed{\,2(x^2+y^2)=z^2\,}\]