The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle.
Let r be the radius, then
OQ = OB = r and OR = (r - 4)
∴ OQ^{2 }= OR^{2 } + RO^{2 }
⇒ r^{2 }= 64 + (r-4)^{2}
⇒ r^{2 } = 64 + r^{2} + 16 - 8r
⇒ 8r = 80 ⇒ r = 10 cm