I'm quoting from the link you posted:
"The bandwidth of the signals either side of the center frequency will be approximately equal to the sampling rate of the PC's sound card. Thus, if the local oscillator is tuned to produce 28.4 MHz to the Quadrature Clock Generators, they will output two signals (I and Q clocks) at 7.1MHz (the "center frequency"). If the PC's sound card has a 48 kHz sampling rate, then the SDR program can translate the QSD's I/Q outputs into a chunk of spectrum that is 24 kHz either side of the center frequency of 7.1 MHz: i.e, 7.076 - 7.124 MHz"
If want to receive 10 MHz you have to replace the 28.4 MHz local oscillator with a 40 MHz oscillator (after :4 you get 10 MHz, +/- 24 kHz).