= Solution
Use target signature $(-,+,\ldots,+)$ and worldsheet signature $(-,+)$. The <Polyakov action> has local <worldsheet diffeomorphism> and <Weyl invariance>, together with global target-space Poincaré symmetry. Under a worldsheet coordinate change $\sigma\mapsto\sigma'$, the embedding coordinates are scalars and the <worldsheet metric> is a tensor:
$$
X'^\mu(\sigma')=X^\mu(\sigma),\qquad
g'_{ab}(\sigma')=\frac{\partial\sigma^c}{\partial\sigma'^a}\frac{\partial\sigma^d}{\partial\sigma'^b}g_{cd}(\sigma).
$$
A <Weyl transformation> acts as $g_{ab}\mapsto e^{2\rho(\sigma)}g_{ab}$ and leaves $X$ unchanged. In two dimensions $\sqrt{-g}$ scales by $e^{2\rho}$ and $g^{ab}$ by $e^{-2\rho}$, so the action is invariant. Target translations act by $X^\mu\mapsto X^\mu+a^\mu$, and <Lorentz transformations> by $X^\mu\mapsto\Lambda^\mu{}_{\nu}X^\nu$ with $\Lambda^{\mathsf T}\eta\Lambda=\eta$; they leave $g$ unchanged. Infinitesimally the latter have $\delta X^\mu=\omega^\mu{}_{\nu}X^\nu$, where $\omega_{\mu\nu}=-\omega_{\nu\mu}$.
Locally, the two diffeomorphism functions put the metric into the form $g_{ab}=e^{2\rho}\eta_{ab}$, and <Weyl invariance> removes the conformal factor. In this <conformal gauge>, the embedding coordinates are free <scalar fields> and satisfy
$$
\boxed{(\partial_\tau^2-\partial_\sigma^2)X^\mu=0.}
$$
The metric equation must still be imposed after <gauge fixing>. Varying the ungauged metric gives the <worldsheet stress-energy tensor> constraint
$$
\partial_aX\cdot\partial_bX-\tfrac12g_{ab}g^{cd}\partial_cX\cdot\partial_dX=0.
$$
With a dot and prime denoting the two coordinate derivatives, the independent <Virasoro constraints> are
$$
\boxed{\dot X^2+X'^2=0,\qquad\dot X\cdot X'=0,}
$$
equivalently $(\dot X\pm X')^2=0$. Local <conformal gauge> leaves residual conformal transformations; on general worldsheet topologies it also leaves global moduli, rather than removing all metric information.
For target translations, differentiation of the Lagrangian with respect to $\partial_aX_\mu$ gives the <Noether current>
$$
j^{a\mu}=-\frac{1}{2\pi\alpha'}\sqrt{-g}\,g^{ab}\partial_bX^\mu.
$$
The embedding equation is $\partial_aj^{a\mu}=0$. In <conformal gauge>, $j^{\tau\mu}=\dot X^\mu/(2\pi\alpha')$, so integration over the <closed string> gives
$$
\boxed{P^\mu=\frac{1}{2\pi\alpha'}\int_0^{2\pi}\dot X^\mu\,d\sigma.}
$$
Periodic boundary conditions make its time derivative vanish. For an infinitesimal <Lorentz transformation>, the antisymmetric coefficient of $\omega_{\mu\nu}$ gives the current $j^{a,\mu\nu}=X^\mu j^{a\nu}-X^\nu j^{a\mu}$. Its divergence vanishes because the remaining product of embedding derivatives is symmetric in $\mu,\nu$. Thus the <Noether charge> is
$$
\boxed{J^{\mu\nu}=\frac{1}{2\pi\alpha'}\int_0^{2\pi}(X^\mu\dot X^\nu-X^\nu\dot X^\mu)\,d\sigma.}
$$
These are the <target-space Noether charges of a string>.
For the circular solution take $R>0$, and let $\boldsymbol e_r=(\cos\sigma,\sin\sigma)$ and $\boldsymbol e_\sigma=(-\sin\sigma,\cos\sigma)$. Its derivatives are
$$
\dot X=(R,-R\sin\tau\,\boldsymbol e_r),\qquad
X'=(0,R\cos\tau\,\boldsymbol e_\sigma).
$$
Each spatial coordinate has equal second derivatives in $\tau$ and $\sigma$, while the time coordinate has both second derivatives zero. Furthermore
$$
\dot X^2=-R^2\cos^2\tau,\qquad X'^2=R^2\cos^2\tau,\qquad\dot X\cdot X'=0,
$$
so both the field equations and the <Virasoro constraints> hold, including the collapse instants.
At target time $t=X^0=R\tau$, the string is a circle in the $X^1X^2$ plane of radius $R|\cos(t/R)|$. It contracts from radius $R$ to zero and re-expands; the negative cosine reverses material labels by a half-turn rather than creating a negative radius. The spatial configuration repeats after target time $\pi R$. This is a <pulsating circular string>. Its induced metric degenerates at collapse, but the conformal-gauge embedding and constraints give its continuation.
The integrals of $\sin\sigma$ and $\cos\sigma$ vanish, so the momentum is
$$
\boxed{P^\mu=(R/\alpha',0,0,\ldots),\qquad M=R/\alpha'.}
$$
The spatial velocity is radial, giving zero angular momentum density in the plane; all boost-charge terms integrate to zero as well. Hence \b[$J^{\mu\nu}=0$ for every pair of indices].
At maximum expansion, all string elements are instantaneously at rest, and the length is $L_0=2\pi R$. Thus the rest mass per unit length, equal to the <string tension>, is
$$
\boxed{\frac{M}{L_0}=\frac{1}{2\pi\alpha'}.}
$$
At other times the instantaneous circumference is $L=2\pi R|\cos\tau|$, and the elements have speed $|\sin\tau|$. Their Lorentz factor is $1/|\cos\tau|$, so the energy remains $TL/\sqrt{1-v^2}=2\pi TR$. The rest-energy density is $T$; the total rest-frame energy divided by the changing circumference is not constant. This distinguishes the requested intrinsic mass density from the larger energy density of moving string elements.
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