Electronic – Working principle of the AD8232 high pass filter

The "trick" you lack may be using the superposition principle to "ignore" (temporarily) the DC source and consider it to be 0 V while studying only the AC source. Using the superposition principle, you can show that V_out consists of a DC component equal to approx. 1V plus an AC component which depends on Vin and the high-pass filter:

Without \$R_1\$ and the DC source, the cutoff frequency is given by : $$ \omega_c = \frac{1}{2\pi R_2 C_1} = 0.15 Hz $$ (This is a typical RC circuit)

Adding \$R_1\$ and the DC source, the circuit changes a bit, since you are adding a resistor parallel to \$R_2\$, the cutoff frequency is now $$ \omega_c = \frac{1}{2\pi (R_2//R_1) C_1} = 0.238 Hz $$ Since \$R_1//R_2\$ is lower than \$R_2\$, you increased the cutoff frequency, but it's still two decades lower than your expected 50 Hz so it shouldn't be an issue.

Here's a bode plot of a simple RC high pass filter's amplitude response - it contains the formula you need to work out the amplitude at every frequency. Use the formula to decide what cut-off frequency you need and whether the attenuation below 50 kHz is suitable: -

If the filter doesn't attenuate quickly enough below 50 kHz then you need to consider a 2nd order filter or, maybe one that is of a much higher order.

Here's a 2nd order high-pass-filter's response at a cut-off of 48.2 kHz and note that it can produce a peak (resonance) if the Q is too high: -

I've chosen a value of resistance such that the Q is unity - notice the little peak in the response - if R is smaller the peak rises. The benefit of using a 2nd order filter is that instead of a 6 dB per octave fall in gain at low frequencies, the rate is 12 dB per octave (40 dB per decade).

Interactive RLC tool