= Solution
Use a steady, isotropic, self-gravitating <dark matter halo>, with $V$ its constant one-dimensional <velocity dispersion>. The <Spherical Jeans equation> and spherical <mass conservation> read
$$
V^2\frac{d\rho}{dr}=-\rho\frac{GM(r)}{r^2},\qquad
\frac{dM}{dr}=4\pi r^2\rho.
$$
For the scale-free outer region, put $\rho=A r^{-p}$. Its enclosed <mass> scales as $r^{3-p}$, so $GM(r)/r$ is constant only if $p=2$; the <Spherical Jeans equation> then gives $2V^2=GM(r)/r=4\pi GA$. Thus the <singular isothermal sphere> has
$$
\boxed{\rho(r)=\frac{V^2}{2\pi Gr^2},\qquad M(<r)=\frac{2V^2r}{G}.}
$$
Its <circular speed> is $v_c=\sqrt2V$. If a speed denoted $V$ were instead the <circular speed>, the coefficient would be $V^2/(4\pi G)$, but the printed variable is a <velocity dispersion>.
The assumptions matter. A constant <velocity dispersion> and spherical shape alone do not specify a unique central <mass density> or orbital distribution. Even with isotropy, the <Spherical Jeans equation> integrates to $\rho=\rho_0\exp[-(\Phi-\Phi_0)/V^2]$, and the self-gravitating <Poisson equation for Newtonian gravity> gives
$$
\frac1{r^2}\frac{d}{dr}\left(r^2\frac{d\ln\rho}{dr}\right)
=-\frac{4\pi G}{V^2}\rho.
$$
There are finite-central-density isothermal solutions as well as the singular one; the $r^{-2}$ expression is the intended outer scale-free halo approximation. A finite halo must also end or steepen at large radius, since the untruncated <singular isothermal sphere> has infinite total <mass>.
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