= Solution
On the <unit> <circle> the two equations reduce to the same condition
$$
x^2(f^2+g^2)=f^2,
$$
since $y^2=1-x^2$. Define $H(x,y)=x^2(f(x,y)^2+g(x,y)^2)-f(x,y)^2$. At either point $(\pm1,0)$ it is strictly positive, and at either point $(0,\pm1)$ it is strictly negative. On each of the four closed quarter-circle arcs, continuity and the <intermediate value theorem> give a zero in its interior. These four interiors are disjoint, so \b[there are at least four distinct solutions], one in each open quadrant. No solution here lies on an axis because $f,g$ are strictly positive.
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