 Solution

At an arbitrary nonzero vacuum write ${\mathbf{\varphi }}_{\ast }=|v|\mathbf{n}$, where ${\mathbf{n}}^T\mathbf{n}=1$. The Hessian matrix of the potential isOn the vacuum circle this reduces toIt 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.