Question asks for a the parametric equations for the set of all lines that pass through the point $(4,0,0)$ and are tangent to the curve:
$$x^2+y^2+z^2=2$$
My issue here is that I am not really sure how to approach the problem. The previous question asked for a similar thing but it wasn't as complicated as it was just the equation of a circle and not a sphere. Nevertheless, it was still the same point and it was still outside of the equation of the circle.
I am assuming there will be a general solution for this question since the question implies that there will be more than one line that passes through the point and is tangent to the sphere.
Am I supposed to utilize gradients to solve this? Or maybe the 3D tangent line equation? I am genuinely not sure how to approach this problem.
Any help would be appreciated.
Best Answer
This exercise requires no calculus, just basic algebra, trigonometry and analytic geometry.
First let us look at the problem in a general setting with a sphere of radius $r$ centered at the origin and a point $A(a,0,0)$ on the $x$-axis with $a>r$.
The diagram shows a the intersection of the sphere and the plane $z=0$ showing a tangent line to the sphere lying in that plane and intersecting at the point $B$ and intersecting the $y$-axis at point $P$. The coordinates of the various points can be found using basic principles such as ratios in similar triangles and the Pythagorean theorem. You should attempt to verify these for yourself.
The next diagram shows a circle in the plane $x=0$ containing a general point $P$ not necessarily on the $y$-axis as in the first figure, but lying on a circle centered at the origin and having radius $R=\frac{ar}{\sqrt{a^2-r^2}}$.
With this setup we can now express a general tangent line to the sphere $x^2+y^2+z^2=r^2$ as $$ T=A(1-t)+ Pt$$
where $A=(a,0,0)$ and $P=\left(0,\frac{ar}{\sqrt{a^2-r^2}}\sin\theta,\frac{ar}{\sqrt{a^2-r^2}}\cos\theta\right)$
So the parametric equations, with parameters $t$ and $\theta$ are
\begin{eqnarray} x&=&a(1-t)\\ y&=&\frac{ar}{\sqrt{a^2-r^2}}t\sin\theta\\ z&=&\frac{ar}{\sqrt{a^2-r^2}}t\cos\theta \end{eqnarray}
In your particular exercise, substitute $r=\sqrt{2}$ and $a=4$.