Find the equation of a cone with vertex at the origin and that passes through the circle $(x-2)^2+(y-3)^2 = 5, z = 3$. 
Find the equation of a cone with vertex at the origin and that passes through the circle $(x-2)^2+(y-3)^2 = 5, z = 3$.

What I've tried:
Whether or not the cone is right circular has not been mentioned explicitly - and we are expected to find the equation of the surface. Next, I am not able to work without the cone's semi-vertical angle. I can find the distance between the origin and (2,3,3), which is the circle's centre; and also the circle's radius, which is the square root of 5. I'm trying to assume a point (x,y,z) on the cone and set up an equation to relate the variables.
I really don't know what do next. Please help me figure this problem out, thanks! (If possible, help me visualise it as well)
 A: Observe that it is a slant cone with circular $xy$-intersection. Because the cone has its vertex at the origin, the dimension of the $xy$-circle, i.e. its center coordinates and radius, changes linearly with $z$.
Given that the center of the $xy$-circle is (2,3) at $z=3$ and due to proportionality with respect to the origin, the center of the $xy$-circle at $z$ is $(2z/3, z)$. Similar, the radius of the circle at $z$ is $\sqrt{5}z/3$.
Therefore, the equation for the cone is, 
$$\left(x-\frac{2}{3}z\right)^2 + (y-z)^2 = \frac{5}{9}z^2$$
A: The desired cone is the union of the lines through the origin $(0,0,0)$ and the points 
$$(2+\sqrt{5}\cos(\theta),3+\sqrt{5}\sin(\theta),3)\quad \text{with $\theta\in [0,2\pi)$}$$ 
along the given circle. Any such line is given in a parametric way as
$$\begin{cases}x(t)=(2+\sqrt{5}\cos(\theta))t\\
y(t)=(3+\sqrt{5}\sin(\theta))t\\z(t)=3t\\\end{cases}$$
Now we eliminate $t$ and $\theta$ and we get
$$(x-2(z/3))^2+(y-3(z/3))^2=5(z/3)^2.$$
that is
$$9x^2+9y^2+8z^2-12xz-18yz=0.$$
