How could I go about determining the component values of the described filter? I know what they are, but not how to get them using proper formulas
I did not realize this circuit would be so complicated to analyse : ) when I started to look at it using the FACTs. First off, I redrawn the circuit in which the inputs were no longer floating but ground-referenced. I also considered the open-loop gain \$A_{OL}\$ to later push it to infinity. The newly-redrawn circuit is below. I have split the potentiometer in two distinct resistances linked by a ratio \$k\$ and reflected in the ground-referenced source driving the op amp:



Both dc and ac responses are similar. The next step is to find the dc transfer function for \$s=0\$: open the capacitors and determine the gain in this configuration:



If you do the maths ok, you should get \$H_0=-\frac{R_3(1-k_1)+R_7}{R_1+k_1R_3}\$

For the next steps, I will determine the time constants \$\tau\$ involving capacitors \$C_1\$ and \$C_3\$. You need to reduce the excitation voltage to 0 V and replace it by a short circuit. You obtain \$\tau_1\$ and \$\tau_3\$. Final step, determine the time constant involving \$C_3\$ while \$C_1\$ is replaced by a short circuit. You obtain \$\tau_{13}\$. Assemble these time constants to form the denominator:

\$D(s)=1+s(\tau_1+\tau_3)+s^2\tau_1\tau_{13}=1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2\$ 

Then, calculate and check if the quality factor \$Q\$ is well below 1 to apply the low-\$Q\$ approximation. You can factor \$D\$ as two cascaded poles if \$Q&lt;&lt;1\$: \$D(s)=(1+\frac{s}{\omega_{p1}})(1+\frac{s}{\omega_{p2}})\$ in which \$\omega_{p1}=Q\omega_0\$ and \$\omega_{p2}=\frac{\omega_0}{Q}\$. Unfortunately here, Q is around 0.47 (\$k=0.6\$) so the poles are not far from each other and the low-\$Q\$ approximation does not hold. The resonant frequency in \$D(s)\$ changes from 492 Hz (\$k=0.1\$) to 1160 Hz (\$k=0.9\$).

The numerator can be determined using several approaches: inspect the transformed circuit and see if any impedance combination could lead to a null in \$V_{out}\$ despite the presence of \$V_{in}\$. You could for example check the condition for which \$\epsilon=0\$ and solve the equation. The other option is to apply an Null Double Injection (NDI) and determine the new time constants (as we did for \$D(s)\$) but with \$V_{in}\$ back in place and while \$V_{out}=0\$. These two paths give the simplest expression for \$N(s)\$. I have adopted the generalized form in which I will calculate three gains \$H\$ when the capacitors are alternatively replaced by a short circuit (for \$a_1\$) or when both are shorted (\$a_2\$). This gives a slightly more complex expression than the two other approaches and you will need to rework the final result to simplify it. For a dynamic response analysis like in here, the generalized formula is okay:

\$N(s)=H_0+s(H^1\tau_1+H^2\tau_2)+s^2H^{13}\tau_1\tau_{13}\$

You can also rearrange this expression in a second-order polynomial form as we did for the denominator. Then assemble the transfer function as \$H(s)=H_0\frac{N(s)}{D(s)}\$. The expressions are given in the below Mathcad files:





I have tweaked these equations so that they depend on \$k\$ as a variable to pass and obtained the following responses:

Magnitude:



Phase:



I did not do it but I believe you can obtain the frequency at which all the curves cross by differentiating the magnitude expression with respect to \$k\$ and solve the value of \$\omega\$ which cancels the expression. You actually calculate the function sensitivity to \$k\$ and check the frequency at which it is 0.

The cool thing with the FACTs is that you do not need to really bother about the circuitry itself: determine the time constants in certain conditions (\$V_{in}=0\$ for \$D\$ and (\$V_{out}=0\$ for \$N\$) and assemble them as recommended to obtain the final expression. If you are interested by the technique, you can check a seminar taught at APEC in 2016. It is an introduction to show the power of the FACTs and to apply them to simple passive and active circuits.