Cone equations given its vertex and a sphere circumscribe in it

Question

In a solved problem it appears the following equations describing the points in a cone with vertex $v=(a,b,c)$ wich circumscribe the sphere $x^2+y^2+z^2=1$ : $$\mathcal{C}: (ax+by+cz-1)^2=(x^2+y^2+z^2-1)(a^2+b^2+c^2-1)$$ I want to know where this equation comes from.

Answer 1

Let:

$$X_1^T=(a,b,c,1) \ \ \text{and} \ \ \ X^T=(x,y,z,1)$$

(homogeneous coordinates).

The given equation has the classical form

$$T^2=SS_1 \ \iff \ (X^TAX_1)^2 = (X^TAX)(X_1^TAX_1)\tag{1}$$

of a cone with vertex $(a,b,c)$ circumscribed to any quadric surface with equation $X^TAX=0$ where $A$ is the associated $4 \times 4$ matrix (here $A=diag(1,1,1,-1)$).

Here is a simple proof:

1. Evidently, $X=X_1$ verifies (1).

2. Consider a ray $X_1+\lambda U$ issued from $X_1$.

In a first step, saying that this ray is tangent to the quadric amounts it has a single common point with it, i.e., the following quadratic equation in $\lambda$ has a double root:

$$(X_1+\lambda U)^TA(X_1+\lambda U)=0$$

$$\lambda^2U^TAU+2 \lambda U^TAX_1+X_1^TAX_1=0$$

Otherwise said, its discriminant must be $0$:

$$(U^TAX_1)^2-(U^TAU)(X_1^TAX_1)=0 \tag{2}$$

Setting

$$p:=X_1^TAX_1, \ \ \ q:=U^TAX_1 \ \ \ \text{and} \ \ \ r:=U^TAU,\tag{3}$$

relationship (2) can be written:

$$q^2=pr \tag{4}$$

Now, in the second step, let us prove that, with $U$ as above, any

$$X=X_1+\mu U, \tag{5}$$

for any $\mu$ (this is the subtle part: taking a general $\mu$, independent of the $\lambda$ above...) verifies equation (1).

Indeed, plugging (5) into (1) gives:

$$((X_1+\mu U)^TAX_1)^2 = ((X_1+\mu U)^TA(X_1+\mu U))(X_1^TAX_1)$$

which, by expansion, using notations (3), becomes

$$(p+\mu q)^2 = (p+2q\mu+r\mu^2)p \tag{6}$$

Taking (4) into account, (6) appears as an identity, valid for any $\mu$, proving that the whole ray $X_1+\mu U$ belongs to the locus.

Remark: A proof for the 2D equivalent expression can be found here.