Band pass (10 kHz, 4 kHz) from normalized LPF TF

Use the low-pass to band-pass transformation $s \to \dfrac{s^2+\omega_0^2}{Bs}$ on the normalized prototype $T(s) = \dfrac{1}{s^2+1.414s+1}$ (second-order Butterworth).

Specifications in rad/s

$$\begin{aligned} \omega_0 &= 2\pi\times10^4 = 62\,831.85\ \text{rad/s}, \quad \omega_0^2 = 3.9478\times10^9 \\ B &= 2\pi\times4000 = 25\,132.74\ \text{rad/s}, \quad B^2 = 6.3165\times10^8 \end{aligned}$$

Substitution

$$T_{BP}(s) = \frac{1}{\left(\dfrac{s^2+\omega_0^2}{Bs}\right)^2 + 1.414\dfrac{s^2+\omega_0^2}{Bs} + 1} = \frac{B^2s^2}{(s^2+\omega_0^2)^2 + 1.414Bs(s^2+\omega_0^2) + B^2s^2}$$

Expanding the denominator:

$$s^4 + 1.414B\,s^3 + (2\omega_0^2 + B^2)s^2 + 1.414B\omega_0^2\,s + \omega_0^4$$ $$T_{BP}(s) = \frac{6.3165\times10^8\, s^2}{s^4 + 3.5538\times10^4 s^3 + 8.5273\times10^9 s^2 + 1.4030\times10^{14}s + 1.5585\times10^{19}}$$

Check: at $s = j\omega_0$, $|T_{BP}| = 1$ (0 dB at 10 kHz), and the 3 dB points are $\Omega_{LP} = \pm1$, i.e. $4$ kHz apart.

Realization as two cascaded biquads

The LP poles $-0.707 \pm j0.707$ each map to a pair of BP poles. Factoring the denominator:

$$T_{BP}(s) = \frac{25\,132.7\,s}{s^2 + 15\,256.8s + 2.9699\times10^9}\cdot\frac{25\,132.7\,s}{s^2 + 20\,280.9s + 5.2479\times10^9}$$

Note $f_{01}f_{02} = (10\text{ kHz})^2$. Each section can be built as a band-pass biquad (e.g. MFB or Tow-Thomas) and the two are cascaded with buffering.

Answer: a fourth-order band-pass filter with the transfer function above, centre frequency 10 kHz, 3 dB bandwidth 4 kHz.