Image taken from maplesoft.com
If you look at the Ebers-Moll model, you can see that each junction is desctibed by a diode and a voltage controlled source. So the collector current will be, as in the first equation, dependent on Vbc (not Vbe) with the Shockley equation, but also on Ib, like in the second equation.
The second factor will be of some meaning only if Vb > Vc, which happens in the saturation region, where the B-C junction is forward biased and current can flow from the base to the collector. In the forward-active (also called linear) region, the B-C junction is reverse biased and the collector current is basically only given by Beta.
In general, it is the states of the PN junctions inside the transistor which will determine what operation region it is in. However, after gathering some experience, one can deduce the states of the above junctions by inspecting the circuit itself without actually measuring the voltages at the terminals.
An example:
Lets analyze the circuit you've referenced.
Once the switch is closed a current of approximately \$1mA\$ will flow into the base, which will cause:
$$V_{BE} \approx 2V$$
Since this is higher than the minimum of \$0.6V-0.7V\$ for being out of cut-off - the transistor is in one of its operational modes. In reality, the Base-to-Emitter voltage will not rise much beyond \$0.6V-0.7V\$ (due to presence of protection resistor R1), which means that the Base current will be a bit higher than \$1mA\$.
Knowing that the motor is \$12V, 100mA\$, and that the transistor is capable of handling \$100mA\$ Collector-to-Emitter current, we can deduce that:
$$I_C = I_{Motor} \approx 100mA$$
Given that we know (from motor's specs) that the motor will consume \$100 mA\$ at \$12V\$, the voltage on the motor:
$$V_{Motor} \approx 12V$$
Which leads to:
$$V_C \approx 0V$$
But this means that Collector-to-Base junction is forward biased which implies that the transistor in saturation.
The above analysis is quite general for this configuration (full voltage rated motor switched by matching BJT), therefore, in majority of circuits like this one, the transistor will be in saturation.
Experienced engineers perform the analysis above at a glance, knowing that the transistor in saturation a second after they see the schematics.
Best Answer
Drive enough current into the base so that the base-collector junction becomes forward biased. How much current will depend on the type of transistor. 'saturation' has to do with how many of the charge carriers in the base region can make it into the collector region. Some will come from the base terminal, but many more will come into the base region from the emitter region. Beyond a certain amount of base current, there just won't be an increase in the available charge carriers that can cross the B-C junction.