Call the other intersection of NP with the circle point Q. Then triangle QXP is similar to triangle XNP. This means
.. QP/XP = XP/NP
.. QP = (XP)^2/NP
By the Pythagorean theorem,
.. XP^2 = XN^2 + NP^2
.. XN = XY/2 = 8 … given in the problem statement
.. QP = (8^2 + 2^2)/2 = 34
The radius is half of QP, so is 17 cm.
