How to calculate the orbital velocity, the high-orbit (quasigeostationary) holding force and the shadow of the disk on Earth?
Let's just pick up the second point. The disk is 42000 km (more or less) from the centre of the Earth, do the gravitational acceleration is $$9.81\times \left(\frac{6371}{41922}\right)^2 =  0.22 ms^{-2}$$ The component of that along the Earth's axis (Southwards) is $$0.22\times\frac{4505}{41922} = 0.024 ms^{-2}$$ 

The mass of the disk is $$5000^2\times\pi\times 4.5 = 3.53\times 10^8 kg$$ so the total force which must be counteracted by thrust is $$3.53\times 10^8\times 0.024 = 8.5\times 10^6 N$$

This is roughly the thrust of 4 of SpaceX's new Raptor engines burning continuously, each consuming about 1 ton per second of propellants. The best ion engines I can see quickly on wikipedia produce about $2\times 10^{-5}
 N/W$ (thrust/power consumption), so you'd need over $10^{12} W$ of power to achieve this thrust with  ion engines. The disk collects at most about 200GW of incoming sunlight, so the numbers don't add up that way. You'd also need quite a lot of Xenon and the ion engines would add a lot of mass.

Moving on to the shadow. Seen from Earth's surface the disk subtends an angle of about $\frac{10}{41922} = 2.5\times 10^{-4}$ radians. The Sun subtends an angle of about $0.01$ radians (half a degree). So the disk would not come close to blocking the Sun at all. That is there would be no umbra at all. In the penumbra the amount of sunlight reaching the Earth would be reduced by about 0.1%