Given:

BC = 6 cm

AB = 8 cm

Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 3: Sum of all angles of a quadrilateral = 360°.

By property 1,

AP = AQ (Tangent from A)

BP = BR (Tangent from B)

CR = CQ (Tangent from C)

∵ ABC is a right-angled triangle, ∴ by Pythagoras Theorem

AC² = AB² + BC²

⇒ AC² = 8² + 6²

⇒ AC² = 64 + 36

⇒ AC² = 100

⇒ AC = √100

⇒ AC = 10 cm

Clearly,

AQ + QC = AC = 10 cm

⇒ AP + RC = 10 cm [∵ AQ = AP and QC = RC]

Also,

AB + BC = 8 cm + 6 cm = 14 cm

⇒ AP + PB + BR + RC = 14 cm [∵ AB = AP + PB and BC = BR + RC]

⇒ AP + RC + PB + BR = 14 cm

⇒ 10 cm + BR + BR = 14 cm [∵ AP + RC = 10 cm and PB = BR]

⇒ 10 cm + 2BR = 14 cm

⇒ 2BR = 14 cm – 10 cm = 4 cm

⇒ BR = 2 cm

Now,

∠BPO = 90° [By property 3]

∠BRO = 90° [By property 3]

∠PBM = 90° [Given]

Now by property 2,

∠BPO + ∠BRO + ∠PBM + ∠ROP = 360°

⇒ ∠ROP = 360° - ∠BPO + ∠BRO + ∠PBM

⇒ ∠ROP = 360° - (90° + 90° + 90°)

⇒ ∠ROP = 360° - 270°

⇒ ∠ROP = 90°

Now, ∵ ∠ROP = 90° and BP = BR which are adjacent sides

∴ Quadrilateral PBRO is a square

⇒ PO = BR = 2 cm

Hence, Radius = 2 cm