First we construct the red circle, whose diameter is $BC$.
Since $N$ is the midpoint of $BC$, it is the center of the red circle.
Since it is given that $\angle BOC=\frac\pi2$, $O$ lies on the red circle.

Next we construct the green circle, whose diameter is $ON$.
Since $ON$ is a radius of the red circle, the green circle is tangent
to the red circle at $O$, with half its diameter. That is, the green circle
is the contraction of the red circle, shrunken by half toward $O$.
This means that $P$, the intersection of the green circle and $OC$, is