In a circle of radius 2 units, a diameter AB intersects a chord of length 2 units perpendicularly at P. If AP > BP, then AP is equal to

Centre of circle = O

PM = PN = 1 unit

OM = 2 units

$\therefore \mathrm{OP}=\sqrt{\mathrm{OM}^{2}-\mathrm{MP}^{2}}$

$=\sqrt{2^{2}-1^{2}}=\sqrt{4-1}=\sqrt{3}$ units

$\therefore \mathrm{AP}=\mathrm{OA}+\mathrm{OP}=(2+\sqrt{3})$ units