Interestingly, if we allow to also specify the sum of the outer face, then it is possible with just 4 circles. The complete graph K4 is planar so it can be drawn, and it has 4 triangular faces, resulting in the matrix
Code:
1 1 1 0
1 1 0 1
1 0 1 1
0 1 1 1
which is invertible. A puzzle based on this might look like this: