Two circles with centres A and B of radii 5 cm and 3 cm respectively touch each other internally. If the perpendicular bisector of AB meets the bigger circle at P and Q, then the value of PQ is

Answer: Option D

According to question

AP = 5 cm, DB = 3 cm
AB = 5 - 3 = 2 cm
AO = AB ÷ 2 = 1 cm
PQ is ⊥ bisector
∴ AO = 1, PO = OQ
In right angle ΔPOA
(AP)² = (OA)² + (OP)²
(5)² = (1)² + (OP)²
(OP)² = 25 - 1
(OP)² = 24
(OP) = 2√6 cm
∴ PQ = 2 × OP
PQ = 2 × 2√6
PQ = 4√6 cm