Each circle in the following order: center, then 1 o'clock, then clockwise.
Circles in the same order: center, then 1 o'clock, then clockwise.
547126
324561
416753
762315
173642
251437
The inference rules were the very basic inference rules of Sudoku: draw the imaginary lines from the existing digits, see if this leaves just one place for a digit in a circle.
For example: from top 1 down left and from the bottom one up left only leaves one spot for a 1 in the 10 o'clock circle - the 3 o'clock position.
The second rule - if there are only two spots for a number in a circle and they are on a line, then you can still draw the imaginary line where this number can't be.
The last rule is simple exclusion - if there are only 2 or 3 free spots in a circle on in a line, see which numbers are missing in this circle, and check the lines that point a those spots against other circles. More often than not, you'll find that for one of the free sports, two of the remaining numbers are excluded.
Those rules are sufficient for the whole puzzle. They come straight for Sudoku and are topology agnostic. I have to say, a Sudoku puzzle than can be solved with those three inferences alone would be considered rather basic.
EDIT with a complete solve.
First, let's introduce some notation. I'll call circles (in the order above) X (center), A (1 o'clock), B, C, D, E, F and the positions within a circle - 0...6, where 0 is the center.
Think imaginary lines from a couple of filled cells in the indicated direction, and how they cover the circle where they intersect.
1@F6↙️ and 1@C4↖️ gives us 1@E2
4@A2↙️ and 4@D4➡️ => 4@C6
2@A1⬅️ and 2@D5↗️ => 2@F3
7@E5↘️ and 7@B3⬅️ => 7@D1
4@C6↖️↗️ and 4@D5↗️ => 4@X1
Exclusion in E => 4@E3
4@X1↘️ and 4@E3➡️ => 4@B0
3@C3↗️ and 3@F1↘️ => 3@B5
1@C4↗️ and 1@F6↘️ => 1@B1
2@D5➡️ => 2@B6
Exclusion in B => 5@B4
5@B4↖️ and 5@X0↖️ => 5@F2
1@B1↖️ and 1@E2➡️ => 1@A5
1@A5↙️ and 1@E2↙️ => 1@D0
5@X0↙️ => 5@D6
6@E6↘️ => 6@D3
Exclusion in D => 3@D2
Exclusion in the horizontal line at C1 =>6@C1
2@F3⬅️ and 2@B6⬅️ => 2@E0
5@D6↖️ => 5@E1
Exclusion in E => 3@E4
3@E4➡️ and 3@B5⬅️ => 3@X6
2@B6⬅️ and 2@A1↙️ => 2@X4
1@B1⬅️ => 1@X3
7@D1↗️ => 7@X2
Exclusion in X => 6@X5
7@X2↗️ and 7@E5➡️ => 7@A6
3@B5↗️ and 3@X6➡️ => 3@A0
6@B2↖️ => 6@A4
Exclusion in A => 5@A3
6@C1↖️ => 6@F0
4@E3↗️ => 4@F5
Exclusion in F => 7@F4
2@E0↘️ and 2@B6↗️ => 2@C2
5@A3↙️ => 5@C5
Exclusion in C => 7@C0
Solved. On the second go, I didn't even use the "two in line" rule.