Since PQ and PR are tangents from point P.We know thatThe radius is perpendicular to the tangent at the point of contact.So, OQ ⟂ PQ and OR ⟂ PRThus, ∆ OQP and ∆ ORP are right triangles.Given:$\angle QPR = 90^{\circ}$$OQ = OR = 4\ \mathrm{cm}$PQ = PR (tangents from same external point are equal)Line OP bisects ∠QPR (property of tangents).So,$\angle QPO = 45^{\circ}$In right triangle OQP$\sin 45^{\circ} = \frac{OQ}{OP}$$\frac{1}{\sqrt{2}} = \frac{4}{OP}$$OP = 4\sqrt{2}\,\mathrm{cm}$