If you look at this problem in 2D you have the following parameters at some instant which describe your trajectory (position and velocity) around a celestial body with gravitational parameter $\mu$: radius $r$, radial velocity $\dot{r}$ and angular velocity $\omega$. There are also a few others, but these do not really matter in this problem, due to symmetry.
You can calculate the radius of your periapsis by using conservation of specific angular momentum and orbital energy. $$ \epsilon=\frac{\omega^2r^2+\dot{r}^2}{2}-\frac{\mu}{r}=\frac{\omega_p^2r_p^2}{2}-\frac{\mu}{r_p} $$ $$ h=\omega r^2=\omega_pr_p^2 $$ From here you can derive an expression for the radius of the periapsis $r_p$: $$ \omega_p=\frac{h}{r_p^2} $$ $$ \epsilon=\frac{\left(\frac{h}{r_p^2}\right)^2r_p^2}{2}-\frac{\mu}{r_p}=\frac{h^2}{2r_p^2}-\frac{\mu}{r_p} $$ $$ \epsilon r_p^2=\frac{h^2}{2}-\mu r_p\rightarrow \epsilon r_p^2+\mu r_p-\frac{h^2}{2}=0 $$ $$ r_p=\frac{-\mu\pm\sqrt{\mu^2+2\epsilon h^2}}{2\epsilon} $$ These two solutions correspond with apoapsis and periapsis (the to point in the orbit at which the radial velocity is zero). And you might suspect that the 'plus' solution might correspond with apoapsis and 'minus' with periapsis. However the opposite is true, since for an elliptical orbit $\epsilon$ is negative (it will also hold for trajectories with higher eccentricity), so: $$ r_p=\frac{\sqrt{\mu^2+2\epsilon h^2}-\mu}{2\epsilon}=\frac{\sqrt{\mu^2+\left(\omega^2r^2+\dot{r}^2-\frac{2\mu}{r}\right) \omega^2r^4}-\mu}{\omega^2r^2+\dot{r}^2-\frac{2\mu}{r}} $$ To find what would be the best angle to burn to lower your periapsis the most you could use a fixed amount of $\Delta v$ and see which angle $\phi$ would lower the periapsis the most ($\frac{\delta}{\delta\phi}\Delta r_p=0$): $$ \Delta r_p=\frac{\sqrt{\mu^2+\left(\omega^2r^2+\dot{r}^2-\frac{2\mu}{r}\right) \omega^2r^4}-\mu}{\omega^2r^2+\dot{r}^2-\frac{2\mu}{r}} - \frac{\sqrt{\mu^2+\left((\omega r+\Delta v\sin{\phi})^2+(\dot{r}+\Delta v\cos{\phi})^2-\frac{2\mu}{r}\right) (\omega r+\Delta v\sin{\phi})^2r^2}-\mu}{(\omega r+\Delta v\sin{\phi})^2+(\dot{r}+\Delta v\cos{\phi})^2-\frac{2\mu}{r}} $$ I will leave deriving this equation to you.
But I suspect that lowering the specific angular momentum will have the greatest impact, so burning in the radial direction, especially when are far away.
Edit:
To answer your second question in your (first) comment. Adding the radial part to $\omega^2r^2$ can be done using the fact that the radial velocity component ca be expressed as $v_\theta=\omega r$, so:
$$
v_\theta+\Delta{v_\theta}=\omega r+\Delta{v}\sin{\phi}\rightarrow \left(v_\theta+\Delta{v_\theta}\right)^2=\left(\omega r+\Delta{v}\sin{\phi}\right)^2
$$
However your equation can be simplified to this as well:
$$
\left(\omega+\frac{\Delta{v}\sin{\phi}}{r}\right)^2 r^2=\left(\left(\omega+\frac{\Delta{v}\sin{\phi}}{r}\right) r\right)^2=\left(\omega+\frac{\Delta{v}\sin{\phi}}{r}\right)^2 r^2
$$