Finding the minimum and maximum distance between the origin and the intersection curve between a cone and a sphere

Question

Suppose you have the cone with its vertex at the origin given by

$ r^T Q r = 0 \tag{1}$

where $r=[x,y,z]^T $, and $Q$ is a $3 \times 3$ symmetric indefinite matrix.

And you have the sphere centered at $C$ with a radius of $a$, whose equation is

$ (r - C)^T (r - C) = a^2 \tag{2}$

I'd like to find the minimum and maximum distance between the origin and the intersection curve between the sphere and the cone. This is the question.

My attempt:

Setting up the Lagrange multiplier problem, the objective function is

$ g(r) = r^T r + \lambda_1 (r^T Q r) + \lambda_2 ( (r - C)^T (r - C) - a^2 ) \tag{3} $

Taking the gradient vector of $g$ with respect to $r$, we get

$ \nabla_r g = 2 r + 2 \lambda_1 \ Q r + 2 \lambda_2 \ (r - C) = \mathbf{0} \tag{4}$

By differentiating with respect to $\lambda_1$ and $\lambda_2$, we get two additional equations:

$ r^T Q r = 0 \tag{5}$

$ (r - C)^T (r - C) - a^2 = 0 \tag{6}$

And the task now is to solve the system of equations $(4), (5), (6)$ for $r$.

Edit:

If $u_1$ and $u_2$ are perpendicular to $C$, then

$u_1^T (\nabla_r g) = 2 \bigg( u_1^T r + \lambda_1 \ u_1^T Q r + \lambda_2 \ u_1^T r \bigg) = 0 $

i.e.

$ (1 + \lambda_2) (u_1^T r) + \lambda_1 \ u_1^T Q r = 0 $

Similarly with $u_2$, we will get,

$ (1 + \lambda_2) (u_2^T r) + \lambda_1 \ u_2^T Q r = 0 $

Since this homogenous system has non-trivial solutions, it follows that its determinant must be zero. That is,

$ (u_1^T r) (u_2^T Q r) - (u_2^T ) (u_1^T Q r) = 0 $

which can be re-arranged, and re-written as,

$ r^T ( u_1 u_2^T - u_2 u_1^T ) r = 0 \tag{3} $

Now we have a system of three quadratic equations $(1),(2),(3)$ in three unknowns which are the coordinates of $r$. This can solved numerically, for example with Sage Math.

Answer 1

Hint.

No need for Lagrange multipliers. The solution points are located on the plane which contains $C$ and the cone revolution axis.

The cone can be represented as

$$ \|r-(r\cdot \vec v)\vec v\|=\alpha |r\cdot \vec v|,\ \ \ \|\vec v\|=1 $$

and this plane can be defined as

$$ \Pi\to (\vec v\times C)\cdot r = 0 $$

and

$$ Q = (I_3-(1+\alpha^2)\vec v\vec v^T)c_0 $$

Now the intersection between $Q$ and $\Pi$, gives two lines ($\mathcal{L}_1,\mathcal{L}_2$). The intersection between the sphere $\mathcal{S}$ and the plane $\Pi$ gives a circle $\mathcal{C}$ and finally the solution points are the intersections of ($\mathcal{L}_1,\mathcal{L}_2$) and $\mathcal{C}$.