Generating a 10 volt rail from a High Voltage Supply with unknown Ground Potential
To supply \$200mA\$ at \$10V\$ from \$200V\$, there must necessarily be \$200mA\$ at \$190V\$ somewhere else, if you are converting the voltage by any linear method (like a voltage divider).

$$ 200mA \cdot 190V = 38W $$

That's a big resistor, transistor, or something, with a big heatsink. Not at all hard, but quite big, and warm.

Probably, you will want a non-linear voltage converter, like a buck converter. This will be much more efficient, and won't need to be big with a large heatsink.

If your control circuitry, and everything connected to it, is connected to only your \$10V\$ output, then it doesn't matter if that \$10V\$ is \$200V - 190V\$ relative to ground, or \$5V - -5V \$ relative to ground. After all, "ground" is just where you stick the "ground" symbol in the schematic, and the electrons can't see that.

But, if you are already using a buck converter, you could get one with an isolated output. Typically, these will use a transformer to couple the power from the input to the output, and the feedback will be done through an optoisolator. Thus, the output of the buck converter is floating, meaning, it has no reference to anything on the input side. You could then, if you wanted, connect your control circuit's ground to some other ground, and not be worried in a voltage difference between these grounds making some large current flow and making smoke.

Even if you don't connect it to something else, that floating output has the nice feature that when someone touches it, maybe accidentally, and that someone also happens to be touching some other potential (say, the Earth, which is connected to a lot of stuff), the output voltage will "float" to their potential, and then there won't be any difference, and they won't be shocked.

It's quite likely that any 200V to 10V buck converter you find will be isolated, for safety. So, perhaps by happy accident, you don't have to do anything special to solve both your problems.