The exercise is the following. Determine

$(i)$ the equation for the cone $C$ with vertex $V=(1, 1, 1)$, tangent to the sphere $S$ with equation $x^2+y^2+z^2=1$;

$(ii)$ a system of equations which determines the circle $\Gamma$ given by the intersection of the cone $C$ and the sphere $S$;

$(iii)$ an equation for the cone $C'$ with vertex $(1, 2, 1)$ and directrix $\Gamma$.

My attempt. I found out that the equation for the cone $C$ required by point $(i)$ is $$ (x+y+z-3)^2=2((x-1)^2+(y-1)^2+(z-1)^2). $$ For the point $(ii)$, is $$ \begin{cases} (x+y+z-3)^2=2((x-1)^2+(y-1)^2+(z-1)^2)\\ x^2+y^2+z^2=1 \end{cases} $$ the required system of equations? How can I find the equation of circle $\Gamma$ as the directrix of the cone $C'$?

Thank You

  • $\begingroup$ Are circle $\Gamma$ from ii and directrix $\Gamma$ from iii supposed to be the same locus? If so, it I'm confused? $\endgroup$
    – Narlin
    Mar 12, 2018 at 18:21
  • $\begingroup$ @Narlin Yes, the exercise says to find the equation for a cone $C'$ with vertex $(1, 2, 1)$ and directrix the circle $\Gamma$. Anyway my problem is to find the circle $\Gamma$ from point $(ii)$.. $\endgroup$
    – Jeji
    Mar 12, 2018 at 18:29

1 Answer 1


!I suppose this is what you are looking for? It seemed to me that you already had it!

You have a point on the circle. To get that point, I suppose you used Pythagoras and a tangent line from V to the circle. Let's call the point on the circle P and lets give it coordinates of $P=(2/3,2/3,-1/3)$ These came from your equation, but I could get them independently from $$|P|=1$$ and $$|P-V|=\sqrt{|V|^2-1}$$ and you have to choose some arbitrary point since there are many tangents to the sphere. These equations depend on your sphere being at the origin. Otherwise they have to slightly modified. Anyway, we have P on the circle.

Now we need the plane of the circle. We know that plane is orthogonal to $\mathbf{r}$ where $\mathbf{r}=(1,1,1)/|(1,1,1)|$ So the equation of the plane is $$ \mathbf{r\cdot}\left(\begin{array}{c}x\\ y\\z\end{array}\right)=\mathbf{r\cdot}\left(\begin{array}{c}P_{x}\\P_{y}\\P_{z}\end{array}\right)$$

For the eqn of the circle, we need two orthogonal unit vectors in the plane of the circle and we need the center of the circle. We have the unit vector $\mathbf{r}$ going to V so we have that line as $(x,y,z)^T = V + t\cdot \mathbf{r}$ We can intersect that line with the plane to get a center point of the cone base as $C=(1/3,1/3,1/3)$

To get the 2 orthogonal vectors (you must have already done this), pick any random vector such as s=(-1,2,2). It doesn't matter so long as it isn't $\mathbf{r}$. Next make $\mathbf{v}=(\mathbf{r}\otimes\mathbf{s})/|(\mathbf{r}\otimes\mathbf{s})|.$ Now $\mathbf{v}$ is one orthogonal unit vector to $\mathbf{r}$ . To get the third one, cross $\mathbf{u}=(\mathbf{r}\otimes\mathbf{v})/|(\mathbf{r}\otimes\mathbf{v})|$. Make sure that $\mathbf{u}$ and $\mathbf{v}$ are unit vectors, orthogonal to each other and in the plane of the circle. It matters. The radius will be $R=|C-P|$

Finallly, the general parameterization of a circle in 3d is $$Circle:\quad\left[\begin{array}{c}x\\y\\z\end{array}\right]=\mathbf{C}+R\cdot cos\theta\cdot\mathbf{u}+R\cdot sin\theta\cdot\mathbf{v}$$ where $\theta$ varies from 0 to $2\pi$.


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