Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 11F / c / Solution

On the unit circle the two equations reduce to the same conditionsince $y^2=1-x^2$. Define $H(x,y)=x^2(f(x,y{)}^2+g(x,y{)}^2)-f(x,y{)}^2$. At either point $(\pm 1,0)$ it is strictly positive, and at either point $(0,\pm 1)$ 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 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.