Perhaps a little amplification of what Owen has said will help clarify
the situation.

Suppose we have a twinlead transmission line, one conductor of which is
carrying a current of 2 amps at one point, and the other 3 amps at the
same point. For simplicity, we'll assume that the currents are exactly
out of phase. The common mode current (as we'll define it*) is 3 - 2 = 1
amp. (We can directly subtract them due to the assumption that they're
exactly out of phase; otherwise we'd have to do a vector addition.) So
the line will radiate exactly as though there was a single conductor
carrying one amp. This is an unbalanced, radiating feedline.

Now let's replace the line with a coax line of the same impedance so it
doesn't otherwise alter the system. What we'll find is that the center
conductor will carry 2 amps. The inner surface of the outer conductor,
which is always forced to be equal and opposite, carries 2 amps of
opposite polarity, that is, 2 amps going exactly the opposite direction.
On the outside of the shield is one amp, our common mode current. The
inner and outer shield currents combine at the cable ends to become 3
amps. This line will also radiate just like a single conductor carrying
one amp.

Finally let's look what happens when we use two coax lines with the
shields connected but floating. Suppose the 2 amps is on the center of
coax A and 3 amps on the center of coax B. On the inside of the coax A
shield is 2 amps flowing one way (the direction opposite the current on
the center conductor). On the inside of the coax B shield is 3 amps,
flowing the other way. What happens at the ends of the shield? At each
end, the 2 amps flowing one way will add to the 3 amps the other way
(since they're connected at the ends so there's a path from one to the
other), resulting in a 1 amp current which flows down the outside of the
shield. This radiates just the same as the others, like a one amp
current flowing on a single conductor. Using dual coax has accomplished
nothing.

The way to prevent the feedline, whatever the type, from radiating, is
to force the currents on the two conductors to be equal and opposite.
This can be done by making both the antenna and the tuner symmetrical,
in which case any of the three lines will be balanced and not radiate.
Another way is to use one or more common mode chokes (current baluns)
which will also balance any of the three line types. But just changing
from one type of line to another doesn't do it.

I've simplified this analysis to deal only with constant currents, such
as you'd approximately have with an electrically short transmission
line. But the individual currents maintain the same ratio all along
longer lines, so the same result occurs.

(*) Common mode current is sometimes defined as half the vector sum of
the two conductor currents, rather than simply the sum as done here. If
you use the other definition, you assume that the common mode current is
flowing on each of the two conductors to determine the amount of
radiation you'll get. The end result is the same either way.

Roy Lewallen, W7EL