= Solution
In a <quasi-circular spiral transfer>, a small tangential thrust $F$ supplies power $Fv_c$ while the orbit stays approximately circular. Differentiating $E=-\mu m/(2r)$ and using $v_c=\sqrt{\mu/r}$ gives
$$
\frac Fm=\frac{\sqrt\mu}{2r^{3/2}}\dot r,\qquad\boxed{\Delta v_{\mathrm{slow}}=\int\frac Fm\,dt=\sqrt\mu\left(\frac1{\sqrt{r_0}}-\frac1{\sqrt{r_1}}\right)=v_0-v_1.}
$$
This is the integrated thrust magnitude per unit constant <mass>, not the net vector change of orbital <velocity>.
For comparison, write $u=\sqrt{r_1/r_0}>1$. Adding the two <Hohmann transfer> impulses and subtracting the slow-transfer result yields
$$
\frac{\Delta v_H-\Delta v_{\mathrm{slow}}}{v_0}=\frac{u-1}u\left[\frac{\sqrt2(u+1)}{\sqrt{u^2+1}}-2\right]<0.
$$
The bracket is negative because $(u+1)^2<2(u^2+1)$ for $u>1$. Thus \b[the Hohmann transfer uses less total impulse than the slow circular spiral] for these ideal coplanar transfers. The <Hohmann transfer compared with a slow spiral> does not establish a global optimum over every possible transfer family or propulsion constraint.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2016-hohmann-and-spiral-transfer.png]
{title=Two tangential Hohmann burns versus an adiabatically slow circular spiral; total impulse is lower for the Hohmann transfer}
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