Question

Determine all the triples (a, b, c) of positive real numbers such that the system
$ax+by-cz=0$
$a\sqrt{1-x^2}+b\sqrt{1-y^2}+c\sqrt{1-z^2}=0$, is compatible in the set of real numbers, and then find all its real solutions

Answer 1

Clearly $|x|\le 1$. As x runs over $[-1,1]$, the vector $u=(ax,a\sqrt{1-x^2})$ runs over all vectors of length a in the plane having a non negative vertical component.
Putting $v=(byb\sqrt{1-y^2},w=(cz,c\sqrt{1-z^2})$, the system becomes $u+v=w$, with vectors u, v, w of lengths a, b, c respectively in the upper half-plane. the a, b, c are sides of a (possibly degenerate) triangle; i.e. $|a-b|\le c\le a+b$ is a necessary condition.
Conversely, if a, b, c satisfy this condition, one constructs a triangle OMN with $OM=a,ON=b,MN=c$. If the vectors $\overrightarrow{OM},\overrightarrow{ON}$ have a positive nonnegative component, then so does their sum. For every such triangle, putting $u=\overrightarrow{OM},v=\overrightarrow{ON}$, and $w=\overrightarrow{OM}+\overrightarrow{ON}$ given a solution, and every solution is given by one such triangle. This triangle is uniquely determined up to congruence: $\alpha =angleMON=angle(u,v)$ and $\beta =angle(u,w)$.
Therefore, all solutions of the system are
$x=cost,y=cos(t+\alpha ),z=y=cos(t+\beta ),tϵ[0,\pi -\alpha ]$ or
$x=cost,y=cos(t-\alpha ),z=y=cos(t-\beta ),tϵ[\alpha ,\pi ]$.