2 and 3
The 2 is already circled, so we can grey out all the other 2s in the grid.
There's only one 3 in the grid, so circle it and grey out everything else that follows from the 2 and 3 in their rows/columns.
8 and 9
There are only three ungreyed 8s in the grid now (on the fourth and fifth rows), so one of them must be circled.
There's also two 9s in those rows; assuming either of those 9s circled, we would know exactly which 8 is circled, and we can quickly obtain a contradiction. So both of those 9s can be greyed out.
There are only three ungreyed 9s in the grid now (top left corner), so one of them must be circled.
6 and 7
There are only three ungreyed 6s and three ungreyed 7s in the grid now.
If the 6 in the top row is circled, then the 7 in the fifth row must be circled, and then the 8 in the fourth row must be circled, and we end up with no options for circled 4, contradiction.
If the 6 in the fourth row is circled, then after some greying out we find that the circled 7 must be in the top row and the circled 5 must be in the second row, so there's no options for circled 9, contradiction.
Now we know that
the circled 6 must be in the bottom row, then the circled 8 must be on the right side, then the circled 7 must be in the fourth column, then the circled 4 and 5 must be on the left side ... the deductions fall like dominoes.
Step by step