I can walk you through the steps.
Assuming that Vi is the initial launch velocity, Phi is the angle from horizontal of the Launch and Re is the radius of the Earth:
Then from
(1) [tex]V_i = \sqrt{k \left ( \frac{2}{R_e}- \frac{1}{a} \right )}[/tex]
where k = 3.987e14 m^3/s^2
you can solve for 'a', the semi-major axis of the orbital trajectory.
Then, with
(2)[tex]P = 2 \pi \sqrt {\frac{a^3}{k}}[/tex]
you find P, the period of the orbit.
Then from the relationships:
(3)[tex]A = \frac {V_i R_e \cos \phi}{2}[/tex]
and
(4)[tex]A = \frac{\pi a^2 (1-e^2)}{P}[/tex]
Where A is the Areal velocity, you can solve for 'e' the eccentricity.
Now, from
(9)[tex]R_{apo} = a(1=e)[/tex]
You get the apogee radius
This minus R_e gives you the altitude of apogee.
using equation (1) from above and substituting R_apo for R_e and using the above arrived at value for a, we get the orbital speed at apogee (V_apo)
The circular orbital speed at the apogee altitude is
(10) [tex]V_o = \sqrt{\frac{k}{R_{apo}}}[/tex]
and the difference between this and V__apo is the delta v needed to circularize the orbit at this altitude.
To work out the delta v needed to circularize the orbit at some lower altitude than apogee takes a bit more work.
Assuming that 'r' is the radial distance to the center of the Earth where you want to circularize the orbit, then you need to solve
(11) [tex]r=a \frac{1-e^2}{1+e \cos \theta}[/tex]
for the theta, the angular distance from perigee.
Then with
(12) [tex]\tan f = \frac{e \sin \theta}{1+ e \cos \theta}[/tex]
you solve for 'f' the angle to the horizontal for the object's velocity at that point.
Again equation (1) using 'r' gives the magnitude of the velocity(v) at this point. and equation (10) gives the circular orbital speed at this altitude ( again using 'r')
a little vector addition between the present orbital velocity (speed and direction) and the desired circular orbit velocity gives you the required delta v.
The other way to find 'f' is to note that the Areal velocity at apogee is
(13) [tex]A = \frac{V_{apo} R_{apo}}{2}[/tex]
And since Areal velocity is constant throughout the orbit,
equations (13) and (3) can be equated by using r and v and solving for phi (which in this case will be 'f')
In your original scenario, you gave phi, the initial launch angle, r, the altitude of the object at a point of its trajectory and v, the magnitude of the velocity at that point.
To work out the information you wanted, you do the following:
With r and v solve equation (1) for a
With R_e and a solve for V_i again using equation (1)
With phi and V_i use equation(3) to solve for the Areal velocity
With the Areal velocity, v and r, use equation (3) to solve for phi(f)
Solve for the circular orbital velocity at r using equation (10)
Use vector addition to get the difference between circular and present orbital velocities.
One last point, the reason the total delta v to reach a circular orbit at 400 km was higher with your original set up than the other two was that with the original parameters, you were "overshooting the mark" and the projectile was moving too fast when it got to 400 km and you used up delta v shedding the excess.
Now if you let the projectile continue on its way to its apogee (1298 km altitude) it would have only taken a delta v of 346 m/s to attain a circular orbit at that altitude. This means for a total cost of 8.6 km/sec you get a higher orbit than you would trying the circularize the same launch trajectory at 400 km at a cost of 8.94 km/sec
