Correct Option: B

On the basis of given in question , we draw a figure of a circle with centre O in which T is an external point where OT = 13 cm and OT intersects the circle at point E and AB is a tangent at point E

∠OPT = 90°
In ∆OPT,
OT² = OP² + PT²
⇒ 13² = 5² + PT²
⇒ PT² = 169 – 25 = 144
⇒ PT = √144 = 12 cm.
Tangents drawn from an external point on a circle are equal.
∴ Let AP = AE = y
⇒ AT = PT – AP = (12 – y) cm
OE ⊥ AB
⇒ ∠OEA = 90°
⇒ ∠ AET = 90°
∴ AT² = AE² + ET²
⇒ (12 – y)² = y² + (13 – 5)²
⇒ 144 – 24y + y² = y² + 64
⇒ 24y = 80