This is a:
MINESWOKU puzzle!
Because:
The sudoku is easily solved using only basic deductions:
Now we have a number of circled digits that can be treated as minesweeper clues, allowing us to flag a number of cells.
(Marking mines with grey background, and known safe cells with green)
The 5 in R1C4 must be surrounded by mines.
Then the only way to get a sum of 20 in the first row is by flagging 8 and 2.
In row five, the only way to get a sum of 4 is as 1+3, so the rest of that row is safe.
And we can mark 8 and 9 safe in row six.
In row two, we have a total of 9 in marked mines, so need 21 more from (9,8,7,4,3) which can only be 9,8,4.
The 5 in R2C8 needs two more mines. They cannot be the 9 and the 6, as that would exceed the row three sum of 14.
So we must have the 1 and one of (6,9) as mines.
That means the other three unknowns around R4C7 must all be mines.
And that gives us a 7 in row three, so we know the 6 must be the last mine in this row.
The 4 in R5C6 needs two more mines. We cannot have both the 5 and 6 in row six (row sum of 7), so the 1 in R4C5 must be a mine.
And we only need 1 or 2 more in row six, so everything else is safe.
Row four needs 12 more from (9,8,7,6,3) which can only be 9,3.
The 1 in R6C3 must be a mine to complete the 2 above.
Then the other mine in row six must be the 6.
The 8 in row seven cannot be a mine, because that would make all other cells except 1 in that row safe, which would make the uncircled 2 above a correct clue.
So (still in row seven) to avoid making the 2 in R6C8 a correct clue, either both the 2 and 3 are mines or neither are.
If both are mines, we need 4 to be a mine to complete the row.
If neither are mines, we need the 5 to be a mine to complete the 3 in R6C7, and again need the 4 to complete the row.
So we know the 4 is a mine, with either (2,3) or 5. And the rest of that row is safe.
The 9 in row eight cannot be a mine, because that would make the rest of that row safe, leaving too few possible mines for the 3 in R9C6.
For the same reason, the 6 in that row cannot be a mine, as the only way to complete the row would then be the 3, again making the rest of the row safe.
And, the 7 in that row cannot be a mine, as the 2 is circled.
The 9 in row nine cannot be a mine, because that would make the rest of the row safe, requiring three mines in the row above the 3 in R9C6, which would exceed the total for row eight.
From here, I struggled to find a logical path forward.
If the 8 in row 8 is a mine, then (after a chain of mine placements) the uncircled 3 in R8C2 ends up being a correct clue.
So the 8 must be safe, and the 5 must be a mine.
Now the 1 and 4 in row eight cannot both be mines, so the 7 and 1 in row nine must be for the 3 in R9C6.
In row nine, that leaves us needing 10 more from (8,6,5,4,2) which must be (8,2) or (6,4). So the 5 is safe.
But from there, I can find three apparently valid solutions, so what have I missed?:
The intended solution (applying A1Z26 to the sum of mines in each column to get the solution):
Two other solutions to the minesweeper puzzle:
