Question

a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8cm and RB = 4cm , find the radius of the circle.

Answer 1

given... AB is the diameter and PQ is the chord of the circle with centre O and AB bisects PQ at R also   PR = RQ = 8 cm and RB = 4 cm
we know that the line from the centre of the circle which bisects the chord is perpendicular to the chord
⇒ AB ⊥ PQ
Let the radius of the circle be r
⇒ OP = OB = r
⇒ OR + RB = r
⇒ OR = r – RB = r-4
Now in right ∆ ORP
OR²+PR²=OP²
⇒ (r – 4)²+8²=R²
⇒r²+4²-2*4*r+64=r²
⇒16-8r+64=0
⇒8r=80
⇒r=80/8
⇒r=10
∴RADIUS 10

Answer 2

AB is a chord of circle with centre O as shown in figure.

according to question,

PQ passing through centre O. hence, PQ is a diameter of circle { as we know, a line segment passing through centre cut the circle at two points is known as diameter of circle }

so, PQ is diameter of circle and we know, half of diameter of circle is known as radius of circle. here POQ = 2AB , means Length of AB is equal to length of radius.

Let radius of circle is r

then, AB = r

from figure it is clear that OA and OB are the radius of circle .

so, OA = OB = r