Question

⭐ Hello Users ⭐

Prove that in two concentric circles,the chord of the larger circle ,which touches the smaller circle ,is bisected at the point of contact.

✨Hoping for good answers✨

Answer 1

Let O be the centre of two concentric circles C1 and C2 .
Let AB is the chord of larger circle, C2​, which is a tangent to the smaller circle C1 at point P.
Now, we have to prove that the chord AB is bisected at P , that is AP = PB.
Join OP.
Now, since OP is the radius of the circle C1 and AB is the tangent to C1 at P.
So, OP ⊥ AB    [ tangent at any point of circle is ⊥ to radius at point of contact]

Since AB is the chord of the circle C2 and ​OP ⊥ AB,
⇒ AP = PB   [perpendicular drawn from the centre to the chord always bisects the chord]

Answer 2

Let O be the centre of two concentric circles C1 and C2 .


Let AB is the chord of larger circle, C2​, which is a tangent to the smaller circle C1 at point P.



Now, we have to prove that the chord AB is bisected at P , that is AP = PB.
Join OP.



Now, since OP is the radius of the circle C1 and AB is the tangent to C1 at P.



So, OP ⊥ AB    [ tangent at any point of circle is ⊥ to radius at point of contact]

Since AB is the chord of the circle C2 and ​OP ⊥ AB,


⇒ AP = PB   [perpendicular drawn from the centre to the chord always bisects the chord]


$hope \: helps$