= Solution
Assume the station's <mass> remains fixed through an instantaneous kick, with subsequent motion governed only by the Earth's <Newtonian gravitational field>. Put $\kappa=GM_E$, and work per unit station <mass>. The initial <circular orbit> has <specific angular momentum> $j$ and <specific orbital energy> $\varepsilon_0$ given by
$$
v_0^2=\frac{\kappa}{r_o},\qquad
j^2=\kappa r_o,\qquad
\varepsilon_0=-\frac{\kappa}{2r_o}.
$$
Let $\delta$ denote the fractional increase of <kinetic energy>. The position and <potential energy> are unchanged during the kick, so
$$
\varepsilon=\varepsilon_0+\delta\frac{\kappa}{2r_o}
=-\frac{\kappa(1-\delta)}{2r_o}.
$$
Unchanged <angular momentum> at unchanged position fixes the tangential <velocity>; the added <kinetic energy> must be radial. Thus this is a radial kick, not a purely tangential speed increase. The new <orbital eccentricity> follows from
$$
e^2=1+\frac{2\varepsilon j^2}{\kappa^2}=\delta.
$$
At either radial <turning point>, the <Kepler radial energy equation> gives $\varepsilon=j^2/(2r^2)-\kappa/r$. Its two roots, equivalently the <periapsis> and <apoapsis> of the <elliptic orbit>, are
$$
\boxed{r_p=\frac{r_o}{1+\sqrt\delta},\qquad
r_a=\frac{r_o}{1-\sqrt\delta}.}
$$
For $\delta=0.01$, $e=0.1$ and
$$
\boxed{r_p=\frac{10}{11}r_o,\qquad r_a=\frac{10}{9}r_o.}
$$
An inward or outward kick gives the same pair of radial <turning points>, although it places the station at a different phase of the <orbit>. These two finite positive radii require $0\leq\delta<1$.
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