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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 = PC2+PO2 = QB2+QO2

∴ 32+x2 = 22+(x+1)2

∴ x = 2

∴ Radius = 92+22 = √13

Hence, option (b).

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