Get clean 9VDC power from 9VAC using a LM317T
The Vadj pin on your LM317 is vitally important to the circuit. Without it connected correctly you'll likely have the same noisy DC supply, just 1.5V to 2V lower than whatever is coming out of the bridge recitfier.

The normal method of using a LM317 is much like your diagram, but could use a few less resisors.
The way it operates is to regulate the voltage between its Vout &amp; Vadj pins to 1.25V.
So how does this circuit produce 9V you may ask?
The 1.25V is put across a resistance (in your case R1a &amp; R1b in parallel). Since you now have a voltage across a resistance, there must be current flowing through that resistance - and ohm's law tells you how much (in your case it would be just under 5.2mA).
That current has to flow somewhere and in this circuit it flows down to ground through R2a &amp; R2b. There is also a little bit of current flowing out of the LM317's Vadj pin (it needs some current to run its inernals) and this current also flows down to ground through R2a &amp; R2b. Since there is current flowing through that resistance, ohm's law once again tells us what the voltage across it must be (7.63V).
Now we've worked out that there is 7.63V from ground to the LM317's Vadj pin, and we know that the LM317 regulates the voltage between its Vout and Vadj pins to 1.25V, thereore the voltage at the output is the sum of the two: 8.88V.

How 'Rick Barker' gets to 9.35V is a mystery to me ...
Why he uses sets of parallel resistors (R1a||R1b &amp; R2a||R2b) to achieve such an arbitrary output voltage is also strange. I would probably have used a 240 ohm for R1 and a 1.5k ohm for R2 to achieve an output of 9.14V.

The 10uF 'bypass' capacitor improves the 'ripple rejection' of the circuit (makes it 'cleaner') and the diodes are there to protect the LM317 from any reverse current being forced through it from either the output caps or the bypass cap if the input is somehow shorted to ground.