Question

In a circle of centre O, PR = 3a + 5 and RQ = 5a – 5, OR = 15 units, ∠ORP = 90°. Find the radius of the circle.

• A. 20
• B. 25
• C. 30
• D. 15

Answer 1

Answer:

B. 25

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

⇒ OR bisects the chord PQ.

⇒ PR = RQ

⇒ 3a + 5 = 5a – 5

⇒ a = 5

⇒ PR = RQ = 3 × 5 + 5 = 5 × 5 – 5 = 20

Using Pythagoras theorem,

⇒ OQ² = OR² + RQ²

⇒ OQ² = 15² + 20²

⇒ OQ = 25

Radius of circle is 25 units.