= Solution
At a fixed point with $x,y>0$, the two bracketed factors must vanish:
$$
\mu-y^2-2x^2=0,
\qquad
1-x^2-y^2=0.
$$
Subtracting gives
$$
x^2=\mu-1,
\qquad
y^2=2-\mu.
$$
Hence the interior branch is
$$
\boxed{(x,y)=(\sqrt{\mu-1},\sqrt{2-\mu}),\qquad1<\mu<2.}
$$
Near $\mu=1$, if $\varepsilon=\mu-1$, then
$$
x=\sqrt\varepsilon,\qquad
y-1=\sqrt{1-\varepsilon}-1=-\frac\varepsilon2+O(\varepsilon^2)
=-\frac{x^2}{2}+O(x^4),
$$
which agrees with both the centre manifold and the positive branch of the pitchfork normal form.
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