(d) $\sqrt{2}\,r$
Since AB is tangent to the smaller circle at P, OP ⊥ AB (radius ⊥ tangent). AB = 2r, so PB = r. In right △OPB: $OB^2 = OP^2 + PB^2 = r^2 + r^2 = 2r^2$, giving $OB = r\sqrt{2}$.
Key steps: tangent ⊥ radius (Theorem 10.1) makes △OPB right-angled at P; perpendicular from centre bisects chord, so PB = r; then Pythagoras gives the radius of the larger circle as $\sqrt{2}\,r$. Examiners expect the right-angle reasoning and Pythagoras application clearly stated.