Question

In the following figure, AD is a straight line, OP ⊥ AD and O is the centre of both circles. If OA = 34cm, OB = 20 cm and OP = 16 cm;
find the length of AB.

Answer 1

For the inner circle, BC is a chord and OP ⊥ BC.

We know that the perpendicular to a chord, from the center of a circle, bisects the chord.

∴ BP = PC
By Pythagoras theorem,
OB² = OP² + BP²
⇒ BP² = 20² - 16² = 144
∴ BP = 12 cm

For the outer circle, AD is the chord and OP ⊥ AD.

We know that the perpendicular to a chord, from the center of a circle, bisects the chord.

∴ AP = PD
By Pythagoras Theorem,
OA² = OP² + AP²
⇒ AP² = (34)² - (16)² = 900
⇒ AP = 30 cm

AB = AP - BP = 30 - 12 = 18 cm