A cone with guiding curve $x^2+y^2+2ax+2by=0$ contains $(0,0,c)$. Its section by $y=0$ is a rectangular hyperbola. Prove its vertex lies on a circle. 
A cone has its guiding curve to the circle $x^2+y^2+2ax+2by=0$ and passes through a fixed point $(0,0,c)$. If the section of the cone by plane $y=0$ is a rectangular hyperbola. Prove that the vertex lies on fixed circle $x^2+y^2+z^2+2ax+2by=0$ and $2ax+2by+cz=0$.

Attempt:
Using equation of circle and fixed point, equation of cone can be found out, it comes:
$$cx^2+cy^2-2axz-2byz+2acx+2bcy=0$$
Its section by $y=0$ plane comes out to be
$$cx^2-2axz+2ax=0$$
I am unable to proceed further. Please help
 A: Let $V=(x_0,y_0,z_0)$ be the vertex of the cone. The cone is composed of all lines passing through $V$ and a point of circle $\gamma$ of equation $x^2+y^2+2ax+2by=0$ in the $xy$ plane; the equation of the cone is then:
$$
(z_0x-x_0z)^2+(z_0y-y_0z)^2+2a(z_0x-x_0z)(z_0-z)+2b(z_0y-y_0z)(z_0-z)=0.
$$
Intersecting this with plane $y=0$ gives the equation of a conic in the $xz$ plane:
$$
(z_0x-x_0z)^2+y_0^2z^2+2a(z_0x-x_0z)(z_0-z)-2by_0z(z_0-z)=0.
$$
This represents a rectangular hyperbola if the coefficients of $x^2$ and $z^2$ in that equation are opposite, which leads to the equation:
$$
\tag{1}
x_0^2+y_0^2+z_0^2+2ax_0+2by_0=0.
$$
This is the equation of the sphere having $\gamma$ as a great circle.
We know, on the other hand, that point $P=(0,0,c)$ lies on the cone, implying that $V$ belongs to the cone having $P$ as vertex and $\gamma$ as guiding curve. Hence the coordinates of $V$ must satisfy the equation:
$$
\tag{2}
cx_0^2+cy_0^2+2ax_0(c-z_0)+2by_0(c-z_0)=0.
$$
Combining equations $(1)$ and $(2)$ we obtain the equation of a plane, to which $V$ must then belong:
$$
\tag{3}
2ax_0+2by_0+cz_0=0.
$$
Vertex $V$ must then lie on the intersection of sphere $(1)$ and plane $(3)$, which is exactly the circle we were required to find.

A: That the guiding curve is a circle implies that the axis of the cone is perpendicular to the plane of the circle (that is, the $xy$-plane, which we'll call "the horizontal") and passes through the circle's center, $(-a,-b,0)$. 
Also, it is "known" (and shown in this answer) that the eccentricity of a conic is given by
$$e = \frac{\sin \angle P}{\sin \angle C} \tag{1}$$
where $\angle C$ and $\angle P$ are the "cone angle" and "plane angle", the angles with the horizontal made by the cone's generator-lines and the cutting plane, respectively. Now, a rectangular hyperbola has eccentricity $\sqrt{2}$, and the plane $y=0$ makes $\angle P = 90^\circ$ with the horizontal; we find, then, that $\angle C = 45^\circ$. Consequently, the height of the cone's vertex above (or below) the $xy$-plane must be equal to the guiding circle's radius, $\sqrt{a^2+b^2}$.
The vertex of the cone is thus 
$$V = (-a,-b,\pm\sqrt{a^2+b^2}) \tag{2}$$
which clearly satisfies the target equation
$$x^2 + y^2 + z^2 + 2 a x + 2 b y = 0 \tag{3}$$
for the sphere that has the guiding circle as a great circle.
Note that we have not yet introduced the "fixed point" $C := (0,0,c)$. Since we have deduced the cone's vertex without this point, we find that $c$ is not actually a free parameter in this problem. Indeed, one readily sees that, because $\overline{OV}$ and $\overline{VC}$ are generators making $45^\circ$ angles with the horizontal, $\triangle OVC$ is an isosceles right triangle with apex $V$; necessarily, the $z$-coordinate of $C$ must be twice the $z$-coordinate of $V$: that is, $c = \pm 2 \sqrt{a^2+b^2}$. Thus, the target plane
$$2 a x + 2 b y + c = 0 \qquad\to\qquad a x + b y \pm z \sqrt{a^2+b^2} = 0 \tag{4}$$
is easily seen to be satisfied by $V$. $\square$
