i. seg OP is the radius of the circle.
∴ d(0, P) = 9 cm
ii. Here, 8 cm < 9 cm
∴ d(0, Q) < d(0, P)
∴ d(0, Q) < radius Point Q lies in the interior of the circle.
iii. There can be two locations of point R on line l.
d(0, R) = 15 cm
Now, in ∆OPR, ∠OPR = 90° [Tangent theorem]
∴ OR2 = OP2 + PR2 [Pythagoras theorem]
∴ 152 = 92 + PR2
∴ 225 = 81 + PR2
∴ PR2 = 225 – 81 = 144 [Taking square root of both sides]
∴ PR = √144
∴ PR = 12 cm