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.

$|AB| =\sqrt{(\frac ac - \frac bd)^2 +(\frac1{2c^2} - \frac1{2d^2})^2)} = \frac1{2c^2} + \frac1{2d^2}$

$\Rightarrow \sqrt{\left(\frac{ad-bc}{cd}\right)^2 + \left(\frac{2d^2-2c^2}{4c^2d^2}\right)^2} = \frac{2d^2+2c^2}{4c^2d^2}$

$\Rightarrow \frac{(ad)^2 - 2abcd + (bc)^2}{(cd)^2}+ \frac{4d^4-8(dc)^2 + 4c^4}{16(cd)^4} = \frac{4d^4+8(dc)^2+4c^4}{16(cd)^4}$

$\Rightarrow \frac{(ad)^2 - 2abcd +(bc)^2}{(cd)^2}=\frac{(4d^4 + 8(dc)^2 + 4c^4) - (4d^4 - 8(dc)^2 + 4c^4)}{16(cd)^4}$

$\Rightarrow \frac{(ad-bc)^2}{(cd)^2} = \frac{16(cd)^2}{16(cd)^4}$
$\Rightarrow (ad-bc)^2=1$
$\Rightarrow |ad-bc|=1$.

We also received solutions from Sarith from Royal College in Sri Lanka, and Vignesh from Hymers College in the UK. You can read their solutions below:

Sarith's Solution
Vignesh's Solution