In the given figure, two concentric circles are drawn. If the radius of an outer circle = 13 cm and the length of the chord of an outer circle which touches an inner circle = 24 cm then, find the radius of inner circle.
Given - Radius of outer circle = 13 cm and PQ = 24 cm. PQ is the chord of the outer circle which touches the inner circle at L. Join OL and OP.
By Theorem- In two concentric circles, the chord of the bigger circle, that touches the smaller circle is bisected at the point of contact with the smaller circle.
∴LP=12PQ
∴LP=12×24=12cm
so,
LP = 12 cm and OP = 13 cm (Radius of outer circle)
By Theorem- The tangent at any point of a circle is perpendicular to the radius through the point of contact.
so, ∠OLP = 90∘
In right Δ OLP, applying pythagoras theorem,
OP2=OL2+LP2
⇒(13)2=(OL)2+122
⇒OL2=169−144
⇒OL2=25
⇒OL=√25=5 cm
∴ Radius of inner circle (i.e OL) = 5 cm