Okay, I had said that I would post another answer based on your clarification, but then I got temporarily distracted. Here it goes now.
Here is just what you've asked for; How to calculate the radii of high orbits above the parallels of 45 and 80 degrees and then some even higher ones!
Just fyi I have confirmed by a simple 3D orbit simulation that this works. I calculate the real acceleration, then zero-out the z-component in order to simulate the thrust. Since the magnitude of the thrust is not part of the question, there was no need to calculate it explicitly, but it will be sizable, and a realistic conventional spacecraft couldn't maintain it very long.
We will assume you have some "vertical" acceleration due to thrust $a_T$ produced by some unconventional thrust mechanism pushing your spacecraft "up and away" from the plane of the equator, that exactly cancels the "downward" component of Earth's gravitational acceleration. This leaves only a horizontal acceleration $a_H$.
Every unconventional orbit is defined by the axial distance $R$ and the height above the equatorial plane $h$. The distance to the center of the earth $r=\sqrt{R^2 + h^2}$ is used to calculate the radial acceleration
$$a = \frac{GM}{r^2} = \frac{GM}{R^2 + h^2}$$
and the horizontal acceleration is (by use of similar triangles)
$$a_H = \frac{GM}{R^2 + h^2} \times \frac{R}{\sqrt{R^2 + h^2}} = \frac{GM \ R}{(R^2+h^2)^{3/2}}.$$
For a circular orbit the acceleration is $v^2/r$, so we can write
$$v^2 = a_H R = \frac{GM \ R^2}{(R^2+h^2)^{3/2}}.$$
Also for a circular orbit, the relationship between the radius, velocity and the period is
$$T = \frac{2 \pi R}{v}$$
$$v^2 = \frac{4 \pi^2R^2}{T^2}$$
If we set the two expressions for v^2 equal, we get the surprisingly simple result:
$$R^2 + h^2 = \left( \frac{GM \ T^2}{4 \pi^2} \right)^{2/3}$$
Try it! The standard gravitational parameter $GM$ for the Earth is 3.986+14 m^3/s^2. Put in 86164 seconds for T (one sidereal day, 23h 56m 4s) and start with a normal GEO orbit with $h=0$. You should get 42,164 kilometers (42,164,000 meters).
Now, $h$ is given by
$$h = r_E \sin(lat)$$
and let's just use the average radius of the Earth of 6371 km.
Considering that the GEO orbit is so far from Earth, this is only a tiny variation. Let's do some crazy orbits way above the North Pole as well.
lat(degs) h (km) R (km) v (km/s)
--------- ------ ------- --------
0 0 42,164 3.075
45 4,505 41,922 3.057
80 6,274 41,694 3.040
-- 10,000 40,961 2.987
-- 20,000 37,119 2.707
-- 30,000 29,628 2.161
-- 40,000 13,335 0.972