= Solution
A <Hohmann transfer orbit> connects two coplanar circular orbits using two tangential, effectively instantaneous <velocity> changes. Take outward transfer radii $r_2>r_1$, and write $\mu_g=GM$ for the central body's <gravitational parameter>. The intervening ellipse has <periapsis> $r_1$, <apoapsis> $r_2$, <semimajor axis> $a_t=(r_1+r_2)/2$ and <orbital eccentricity> $e=(r_2-r_1)/(r_1+r_2)$.
The circular speed is $\sqrt{\mu_g/r}$. Conservation of orbital <energy> gives the <vis-viva equation> $v^2=\mu_g(2/r-1/a_t)$. Therefore the transfer speeds at the two impulses are
$$
v_{t1}=\sqrt{\frac{2\mu_g r_2}{r_1(r_1+r_2)}},\qquad
v_{t2}=\sqrt{\frac{2\mu_g r_1}{r_2(r_1+r_2)}}.
$$
The first prograde impulse raises the far side of the orbit; the second prograde impulse raises the near side to circularize at $r_2$. Their magnitudes are
$$
\boxed{\Delta v_1=\sqrt{\frac{\mu_g}{r_1}}\left(\sqrt{\frac{2r_2}{r_1+r_2}}-1\right),
\quad
\Delta v_2=\sqrt{\frac{\mu_g}{r_2}}\left(1-\sqrt{\frac{2r_1}{r_1+r_2}}\right).}
$$
The transfer takes half an ellipse period. <Kepler's third law> gives
$$
\boxed{t_t=\pi\sqrt{\frac{[(r_1+r_2)/2]^3}{\mu_g}}.}
$$
For inward transfer the same ellipse is traversed from <apoapsis> to <periapsis> and both burns are retrograde. A rendezvous also needs the correct initial phase of the target: for outward transfer its initial angular lead is $\pi-\sqrt{\mu_g/r_2^3}\,t_t$ modulo $2\pi$. The method assumes a two-body <gravitational field>, negligible atmospheric <drag> and burns short compared with an orbit. It is an economical two-impulse transfer between these circles; it is not universally the least-$\Delta v$ transfer if additional impulses and very large intermediate radii are permitted.
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