Navjot from Sherborne Qatar School sent us a lovely clear explanation:
Explore the value of ad−bc for the touching circles that you have found. What do you notice?
I noticed that $ad-bc$ was always equal to $\pm 1$. This happened when the pair
of fractions were Farey neighbours.
For example, $\frac01, \frac 11: (0 \times 1) - (1 \times 1) = -1$
Or, $\frac 1{11}, \frac 1{12}: (1 \times 12) - (1 \times 11) =1$
In both of these cases, the two circles were tangent to each other.
Can you prove that for any touching circles in the interactivity above, $|ad−bc|=1$?
So it is given to us that the centre of circle $A$ is $\left(\frac ac , \frac1{2c^2}\right)$ with radius $\frac1{2c^2}$,
and the centre of circle B is $\left(\frac bd, \frac1{2d^2}\right)$ with radius
$\frac1{2d^2}$.
To show $|ad-bc| = 1$, I found the length between the centres of the two circles and equated it to the sum of the radii, which is also the length between the two points.