Given:

QP = 4 cm

OQ = 3 cm

SR = 12 cm

SO’ = 5 cm

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

By above property, ∆OPQ is right-angled at ∠OQP (i.e., ∠OQP = 90°) and ∆O’SR is right-angled at ∠O’SR (i.e., ∠O’SR = 90°).

By Pythagoras theorem in ∆OPQ,

OP² = QP² + OQ²

⇒ OP² = 4² + 3²

⇒ OP² = 16 + 9

⇒ OP² = 25

⇒ OP = √25

⇒ OP = 5 cm

By Pythagoras theorem in ∆O’SR,

O’R² = SR² + O’S²

⇒ O’R² = 12² + 5²

⇒ O’R² = 144 + 25

⇒ O’R² = 169

⇒ O’R = √169

⇒ O’R² = 13 cm

Now,

PR = PO + ON + NO’ + O’R

⇒ PR = 5 cm + 3 cm + 5 cm + 13 cm

⇒ PR = 26 cm

Hence, PR = 26 cm