= Solution
\b[False.] The enclosed-mass formula is a consequence of the <spherical shell theorem>. In a spherical <galaxy>, exterior shells exert no radial force and interior mass acts as though concentrated at the centre, giving $v_c^2=GM(<r)/r$.
A razor-thin axisymmetric disc instead has
$$
\boxed{v_c^2(R)=R\,\partial_R\Phi(R,0)=-R\,\partial_R\psi(R,0).}
$$
Both interior and exterior annuli contribute to that derivative. Its geometry is not determined by $M(<R)$.
A concrete counterexample compares a central point mass $M_0$ plus an exterior spherical shell with the same point mass plus a thin circular ring of mass $m$ and radius $a$. Their enclosed-mass profiles are identical: $M_0$ inside $a$, and $M_0+m$ outside. The <interior gravitational force of an exterior thin ring> follows by expanding its angularly averaged relative potential:
$$
\psi_{\mathrm{ring}}(R)=\frac{Gm}{2\pi}\int_0^{2\pi}\frac{d\varphi}{\sqrt{a^2+R^2-2aR\cos\varphi}}=\frac{Gm}{a}\left[1+\frac{R^2}{4a^2}+O\left(\frac{R^4}{a^4}\right)\right].
$$
For $R\ll a$, the ring's acceleration is outward, $\partial_R\psi_{\mathrm{ring}}=GmR/(2a^3)+\cdots$, whereas the spherical shell's acceleration is zero. With $M_0$ large enough to retain <circular orbits>, the disc-side speed is
$$
\boxed{v_{c,\mathrm{ring}}^2=\frac{GM_0}{R}-\frac{GmR^2}{2a^3}+\cdots\ne\frac{GM_0}{R}=v_{c,\mathrm{sphere}}^2.}
$$
A narrow smooth annulus gives the same distinction. This proves that <enclosed mass does not determine a disc rotation curve>, even when the mass profiles of the spherical and disc models agree.
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