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“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 +-1. This happened when the pair of fractions were Farey neighbours.
Eg: 0/1 , 1/1: (0*1)-(1*1) = -1
Or, 1/11, 1/12: (1*12)-(1*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 (a/c , 1/2c^2) with radius 1/2c^2, and the centre of circle B is (b/d, 1/2d^2) with radius 1/2d^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| =√((a/c - b/d)^2 +(1/2c^2 - 1/2d^2)^2) = 1/2c^2 + 1/2d^2
=> √(((ad-bc)/cd)^2 + ((2d^2 - 2c^2)/4(c^2)(d^2))^2) = (2d^2 + 2c^2)/ 4(c^2)(d^2)

=> ((ad)^2 - 2abcd + (bc)^2)/(cd)^2) + (4d^4 - 8 (dc)^2 + 4c^4)/16(cd)^4) = ( 4d^4 + 8 (dc)^2 + 4c^4)/16(cd)^4)

=> ((ad)^2 - 2abcd + (bc)^2)/(cd)^2) = ((4d^4 + 8 (dc)^2 + 4c^4) - ( 4d^4 - 8 (dc)^2 + 4c^4))/16(cd)^4

=> ((ad)^2 - 2abcd + (bc)^2)/(cd)^2) = 16(cd)^2/16(cd)^4

=> (ad-bc)^2/(cd)^2 = 16(cd)^2/16(cd)^4

=> (ad-bc)^2 = 1
=> |ad-bc| = 1
QED

“Can you prove that, given two such circles which touch the x axis at bd and ac, the circle with centre (a+b/c+d,1/2(c+d)^2) and radius 1/2(c+d)^2 is tangent to both circles?”

We already know from the interactivity that the circles are tangent to each other when ad-bc = 1, so all we need to do to prove this is just plug the new coordinates into the equation we used to answer the previous question. Instead of |AB|, we will consider |AC| and |CB| where C is the new circle between the two original ones.

So we know that the coordinates of the new circle C are (a+b/c+d,1/2(c+d)^2) with radius 1/2(c+d)^2. Let m=a+b and n=c+d, making the coordinates (m/n,1/2(n)^2) with radius 1/2(n)^2.

The first method:

|AB| =√((a/c - m/n)^2 +(1/2c^2 - 1/2n^2)^2) = 1/2c^2 + 1/2n^2
=> √(((an-mc)/cn)^2 + ((2n^2 - 2c^2)/4(c^2)(n^2))^2) = (2n^2 + 2c^2)/ 4(c^2)(n^2)

=> ((an)^2 - 2amcn + (mc)^2)/(cn)^2) + (4n^4 - 8 (nc)^2 + 4c^4)/16(cn)^4) = ( 4n^4 + 8 (nc)^2 + 4c^4)/16(cn)^4)

=> ((an)^2 - 2amcn + (mc)^2)/(cn)^2) = ((4n^4 + 8 (nc)^2 + 4c^4) - ( 4n^4 - 8 (nc)^2 + 4c^4))/16(cn)^4

=> ((an)^2 - 2amcn + (mc)^2)/(cn)^2) = 16(cn)^2/16(cn)^4

=> (an-mc)^2/(cn)^2 = 16(cn)^2/16(cn)^4

=> (an-mc)^2 = 1

=> |an-mc| = 1
Doing the same with the coordinates of the centres of circles B and C will result in the same thing.
In fact, |an-mc| = 1 can be worked on more to give ad-bc=1 since m and n were substitutions for a+b and c+d

a(c+d) - c(a+b) = 1
ac+ad-ac-cb = 1
ad-cb =1

This means that the lengths of the radii of circle A and C are Farey neighbours, thus being tangents to each other.