Answer
There are $~~14~~$ points where waves arrive at the detector exactly out of phase.
Work Step by Step
We can let the horizontal line between the two sources be the x axis and let the midpoint between the sources be $x = 0$
We can draw a vertical axis through the point $x = 0$
At the two points where the circle intersects the vertical axis, the path length from the two sources is equal, so the path length difference is zero.
At the two points where the circle intersects the horizontal axis, the path length difference from the two sources is $1.75~m$ which is $3.5~wavelengths$.
Therefore, in each of the four quadrants, there will be one point on the circle where the path length difference is $0.5~wavelengths, 1.5~wavelengths,$ and $2.5~wavelengths$
Note that the waves are exactly out of phase when the path length difference is $(n~\lambda+0.5 \lambda)$, where $n$ is an integer.
Therefore, there are $~~14~~$ points where waves arrive at the detector exactly out of phase.