You have to think in 3D. Basically, what you want is a sphere with round faces of the same diameter arranged as the faces of a soccer ball.
If I was in your shoes, my first idea would be to take a sphere, intersect it with first an icosahedron and then with a dodecahedron (ordre is not relevant).
There are 3 unknown: the radius of the sphere, the radius of the icosahedron, the radius of the dodecahedron.
Actually only 2 because you can always change the scale of the whole model.
But to get the same diameter for all the circles, the distance from the center to the face of the dodecaedron, must be the same as the distance from center to the face of the icosahedron.
This gives you a relashionship between the radius of the dodecahedron and the radius of the icosahedron.
Most of the tools work with the radius of the circumscribed sphere (that is the distance from the center to the vertices) that I will call
Rv. But what we need is the distance to the face (the radius of the incribed sphere
Rf.
Let's do some maths:
for a
icosahedron

Rv20=0.951a (
a is the side of the icosahedon, but we don't care)

Rf20=0.756a
so
Rv20/Rf20=1.258 (no more
a)
For a
dodecahedron
 Rv12=1.401b
 Rf12=1.114b so
Rv12/Rf12=1.258 (no more
b)... The same value. Coincidence? Probably not.
As we stated that we wanted
Rf20=Rf12, it means that
Rv20=Rv12 (very long computation for a very simple result, indeed).
So if you take one dedecaheron and one icosahedron of same radius (
do they intersect into a regular soccer ball?*), and then take a sphere of the same center and make the sphere grow until you see it appearing through the 32 faces and go on growing it before the circles overlap, you should get the correct radius for the sphere.
Then, make some boolean operations to keep the part of the sphere that is inside the dodecahedron and the part of the resulting shape that is inside the icosahedron (in Sketch Up, I think you just have to make the intersection of the three, and then remove manually the useless part of spheres).
Hope this helps.
EDIT:
*do they intersect into a regular soccer ball?
The answer is no (see the render below). The hexagons are not regular. Each hexagon is surrounded by 3 hexagons and 3 pentagons and each pentagon is surrounded by 5 hexagons. The fact that the hexagons are not regular allows the circle inside this kind of hexagon to have a contact with the circles included into the 3 surrounding regular pentagons (but not the circles of the 3 surrounding hexagons), and each circle included into the regular pentagon will touch all five circles of the surrounding irregular hexagons. This was not the case in the original pattern (no contact). We can bet that this shape illustrates the best packing of 32 identical circles onto a sphere.