Find $a$ so that a line is tangent, secant or external from a sphere

Question

I am given the following problem:

Given the line $$r \{ R = (1,0,a) + \lambda [a \quad a \quad 0]$$ and the sphere $$S \{ 8x^2 + 8y^2 +8z^2 - 16x +24y -8z + 19 = 0$$ find, relating to values of $a$, when the line is external, tangent and secant to the sphere.

I completed the square on the sphere's equation for some clarity

$$(x-1)^2 + \left( y + \frac{3}{2} \right)^2 + \left( z - \frac{1}{2} \right)^2 = \frac{9}{8}$$

which gave me a center $C = \left( 1 , - \frac{3}{2}, \frac{1}{2} \right)$ and a radius $r = \frac{3}{2 \sqrt{2}}$.

To evaluate when a line is external or not to a sphere one must relate the distance of the center of the sphere to the line and check if it is greater (or not) than the radius. And my question is: how can I do that given the fact that the line has two variables?

Is the problem not well-made?

Answer 1

The distance squared, writing $c$ for $\lambda$, from $R = (1,0,a) + c [a \quad a \quad 0] = (1+ca,ca,a) $ to $C = \left( 1 , - \frac{3}{2}, \frac{1}{2} \right) $ is

$\begin{array}\\ d^2 &=(1+ca-1)^2+(ca+\frac32)^2+(a-\frac12)^2\\ &=c^2a^2+c^2a^2+3ca+\frac94+a^2-a+\frac14\\ &=2c^2a^2+3ca+\frac52+a^2-a\\ \end{array} $

For fixed $a$, since $\frac{\partial d^2}{\partial c} =4ca^2+3a $ and $\frac{\partial^2 d^2}{\partial c^2} =4a^2 > 0 $, the minimum, unless $a = 0$, is at $4ca^2+3a = 0$ or $c = -\frac{3}{4a}$ when

$\begin{array}\\ d^2 &=2c^2a^2+3ca+\frac52+a^2-a \\ &=2\frac{9}{16a^2}a^2-3\frac{3}{4a}a+\frac52+a^2-a\\ &=\frac98-\frac{9}{4}+\frac52+a^2-a\\ &=\frac98+\frac14+a^2-a\\ &=\frac98+(a-\frac12)^2\\ \end{array} $

From this, $d^2 \ge \frac98$ and, since $r^2 = \frac98$, it seems that the line is never inside the sphere for any $a$ and is tangent to the sphere when $a = \frac12$.

When $a=0$, the line is just the point $(1, 0, 0)$ with distance $d^2 = \frac52$.

Answer 2

let $u$ be the vector from $(1,0,a)$ to the center of the circle (1,-\frac 32, \frac 12)

$u = (0,-\frac 32, \frac 12-a)$

let $v$ be the direction vector of the line. $(a,a,0)$

$\frac {u\cdot v}{v\cdot v} v$ will give the projection of $u$ onto $v$

$u - \frac {u\cdot v}{v\cdot v} v$ will be orthogonal to $v$ and be the shortest vector from the line to the center of the circle

If the magnitude of this vector equals the radius of the circle:

$\|u - \frac {u\cdot v}{v\cdot v} v\| = \sqrt{\frac 98}$

Then our line will be tangent.

$\|(0,-\frac 32, \frac 12-a) - \frac {-\frac 32 a}{2a^2} (a,a,0)\|\\ \|(0,-\frac 32, \frac 12-a) + (\frac 34,\frac 34,0)\|\\ \|(\frac 34,-\frac 34, \frac 12-a)\|\\ \frac {9}{16} + \frac {9}{16} + (\frac 12 - a)^2 = \frac 98\\ (\frac 12 - a)^2 = 0\\ a = \frac 12$