= Solution
At an arbitrary nonzero vacuum write $\boldsymbol\phi_*=|v|\mathbf n$, where $\mathbf n^T\mathbf n=1$. The <Hessian matrix> of the potential is
$$
\frac{\partial^2V}{\partial\phi_a\partial\phi_b}
=8\lambda\phi_a\phi_b+4\lambda(\boldsymbol\phi^2-v^2)\delta_{ab}.
$$
On the vacuum circle this reduces to
$$
\boxed{M^2=8\lambda v^2\mathbf n\mathbf n^T.}
$$
It has <eigenvalue> $8\lambda v^2$ along $\mathbf n$ and <eigenvalue> zero along the one perpendicular direction. Rotating the vacuum conjugates this matrix by an <orthogonal matrix> and cannot change its <eigenvalues>. Thus every point on the circle has exactly one tangential massless <Goldstone boson>.
Geometrically, motion along the circle costs no potential energy whereas motion normal to it does. The circle's dimension, equivalently $\dim O(2)-\dim O(1)=1$, is independent of its point. Counting the number of real field components would instead give two and would incorrectly count the massive radial direction as a <Goldstone mode>.
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