A variable plane parallel to the plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0$ meets the coordinate axes at $A, B, C$. Find the equation of the cone whose vertex is at the origin and the guiding curve is the circle $ABC$.
I have found out that the equation of the circle $ABC$ is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=k$, $x^{2}+y^{2}+z^{2}-k(ax+by+cz)=0$. ($k$ is a constant and the sphere taken passes through the points $A, B, C, O$.)
Now, given the vertex and guiding curve, I know how to find the equation of the cone using the normal method. After some calculation, I eventually obtained the equation of the cone to be: $$x^{2}+y^{2}+z^{2}-(\frac{x}{a}+\frac{y}{b}+\frac{z}{c})(ax+by+cz)=0$$ However, it seems there is a shortcut to solving this particular question. Substitute the value of $k$ from the plane equation in the sphere equation to make the sphere equation homogenous, and there you have it—the equation of the required cone!
I don't really understand this shortcut. I do know that a quadric cone with its vertex at the origin will be a homogeneous second-degree equation, but I don't get this shortcut. A proper explanation will be of great help!