It's generally the first step to calculate the collector resistor Rc, according to the operating collector current Ic and collector-emitter voltage Vce required.
Rc = Vce / Ic
or
Rc = Vcc / Ic(max)
both result the same for a Class A amplifier.
In the "hyperphysics" example however it is said that "The resistance Rc+Re determines the maximum collector current…" and
Rc + Re = Vcc / Ic(max).
http://hyperphysics.phy-astr.gsu.edu/hbase/Electronic/npnce3.html#c1
I find this site to be a very informative and credible one, so it makes me doubt.
Best Answer
If you have an emitter resistor and a collector resistor then the maximum current that can flow is when the transistor is turned on as hard as it can be. If this is assumed to produce a collector emitter volt drop of zero volts then: -
Ic = \$\dfrac{Vcc}{R_E+R_C}\$
If the transistor saturates only to (say) 50 mV then you have to use a figure for Vcc that is 50 mV lower for calculating Ic.
A class A amplifier is going to drive some form of load so it is important to choose a low enough value of Rc so that the application of the load doesn't overly affect the amplifier's gain. So here's another way of looking at it - you consider what the load is and choose Rc accordingly. Then you assume the quiescent operating voltage is half Vcc (or maybe a tad higher if you factor in the emitter dc volt drop) and this dictates Ic.
So no, the first step you take is not generally dictated by the operating current. More often it is dictated by the external load.