Let $\lambda$ be a circle lying in $S$. Then there is a unique plane $P$ in $\mathbb R^3$ such that $P\cap S = \lambda$. Recall from analytic geometry that $$P = \{(x_1, x_2, x_3) : x_{1}\beta_{1} + x_{2}\beta_{2} + x_{3}\beta_{3} = l\}$$ where ($\beta_{1}, \beta_{2}, \beta_{3}$) is a vector orthogonal to $P$ and $l$ is some real number. It can be assumed that $\beta_{1}^{2} + \beta_{2}^{2} + \beta_{3}^{2} = 1$. Use this information to show that if $\lambda$ contains the point $N$ then its projection on $C$ is a straight line. Otherwise, projects onto a circle in $C$.
I am trying my hands on this question for fun, but I have no clue. Any help would be greatly appreciated.
