10 kHz bandwidth at 10 MHz is very tight for a R-L-C filter. Even if you could put a high enough order filter together, it would be useless due to part tolerance errors.
The only passive way to do this that has any chance of working is to use a 10 MHz crystal. You should still preceed it with a L-C filter to eliminate frequencies that can make the crystal resonate at overtones (harmonics). The L-C pre-filter will also help reduce the power of the signals the crystal has to get rid of.
There is another way, but it is definitely active and more complex, and uses the technique of hetrodyning. The basic concept is to shift the original frequency to a lower value where the desired bandwidth is a much larger fraction of the frequency, then shift the result back. The relatively wider bandwidth at the lower frequency makes a filter more tractable. Old AM radios used this technique, but didn't bother shifting back since they only wanted the amplitude and could get that from the shifted frequency.
450 kHz was a common IF (intermediated frequency) for AM radios intended to receive the commercial AM band from about 550 kHz to 1.7 MHz. The tuning knob would adjust the local oscillator, which needed to be 450 kHz less than the reception frequency. The result would go thru a 450 kHz narrow band filter and amplifier. This needed about 20 kHz bandwidth, which is 4.4% of 450 kHz. That was doable with a few carefully factory-tuned parts. In "super hetrodyne" radios, the tuning knob also adjusted a L-C filter to roughly select the RF frequency of interest. Note that due to how product modulation works (which is how the local oscillator was "mixed" with the filtered RF), there are actually two RF frequencies that result in the 450 kHz IF. These are the local oscillator plus 450 kHz (the desired RF frequency), and the local oscillator minus 450 kHz, called the "image" frequency. The original L-C filter on the RF needed to be tight enough to eliminate the image frequency before the hetrodyning.
You should also consider what you want to do with the final narrow band signal. If you just want to AM detect it, for example, then there may be other ways than starting with a very narrow band filter. It's not worth going into this without more information about what exactly you are trying to do, where this 10 MHz signal is coming from, what kind of modulation you want to detect, how much out of band noise the input signal contains, etc.
The -3dB point is your cutoff frequency. It's just standard practice to define it that way. In order to find what your values should be, I'd go with equal element implementation (it's simpler, and you can correct for gain with a simple gain stage later if you need to). Choose R1=R2, C1=C2, and pick a value for either R or C. This yields the following formula for the cutoff frequency: $$f_0=\frac{1}{2\pi RC}$$
I generally choose a value of C initially, as it's easier to find or make a resistor with a strange value, whereas it's more difficult with capacitors. So, set your cutoff frequency equal to f0, and solve for R.
Here's an example: let's say I want a LPF with f0=250 Hz. I'll choose C to be 0.1 micro and solve for R.
$$250=\frac{1}{2\pi RC} \rightarrow 250=\frac{1}{2\pi R(0.1x10^{-6})}\rightarrow R\approx6400\Omega.$$
From there, all you need to do is implement your circuit. Once you know what your value for R is supposed to be, you can use a dual-channel potentiometer that has the correct resistance within it's range in place of the two resistors (for the above example, something like a 10k ohm potentiometer would do the trick). This will allow you to change your cutoff frequency, since it's based upon both R and C.
Edit: As Matt Young suggested in the comments, adding a resistor in series with the potentiometer will set the maximum cutoff, and prevent shorts. It's an excellent addition to the circuit, and will keep some sanity when adding the potentiometers.
Best Answer
What you are asking is very difficult. The problem is that you are asking for a constant bandwidth, rather than a constant Q (that is, frequency divided by bandwidth). A 50 Hz bandwidth at 50 Hz is easy - 50 Hz at 5 KHz is very narrow. The sort of circuits you've been looking at are hard to do at very narrow bandwidths due to sensitivity, where the exact bandwidth gets very sensitive to the exact (less than 1%) values of the components you use.
Furthermore, narrow filters are usually intended to have a sharp cutoff at all frequencies away from the center. A simple filter of the sort you link to is called a second-order filter, and for frequencies much away from center the cutoff is quite gradual. You can get around this by specifying higher-order filters, but each of these must be adjusted at the same time by the right amount, and the separate responses tend to interact to produce weird frequency responses.
You have not specified how you intend to change the center frequency. Do you want to use a knob on a front panel, or a voltage? If the former, if you're willing to stick to a 2nd-order filter (the simplest kind), you can get get ganged pots, where 2 resistors are connected to one shaft, and this may possibly do you, although I doubt it.
If you want electronic control, and especially if you want narrow filter response such as 50 Hz at 5 KHz, I suggest you look into switched capacitor filters. Linear Technology and TI both produce compact solutions. LT has an app note http://cds.linear.com/docs/en/application-note/an40f.pdf which is a good place to start.
Most importantly, do some more research on exactly what filter responses mean. Do you need Butterworth response, Bessel or Chebyshev? What order filter do you need? Until you understand the implications of your requirements, it will be hard for you to select the right filter.