Discussion

Explanation:

OP⊥ PQ and OQ⊥ AB

In a circle perpendicular to the chord bisects the chord.
∴ CP = PD = 3 and AQ = QB = 2

Let OP = x cm

OC = OB

In ∆OPC and ∆OQB
​​​​​​​∴ Radius = $\sqrt{P{C}^2+P{O}^2}$ = $\sqrt{Q{B}^2+Q{O}^2}$

∴ $\sqrt{3^2+x^2}$ = $\sqrt{2^2+{(x+1)}^2}$

∴ x = 2

∴ Radius = $\sqrt{9^2+2^2}$ = √13

Hence, option (b).

» Your doubt will be displayed only after approval.