Question

answerrrrrrrr........

Answer 1

Let there is a circle having center O

Let AB is the tangent to the smaller circle and chord to the larger circle.

Let P is the point of contact.

Now, draw a perpendicular OP to AB

Now, since AB is the tangent to the smaller circle,

So, ∠OPA = 90

Now, AB is the chord of the larger circle and OP is perpendicular to AB.

Since the perpendicular drawn from the center of the circle to the chord bisect it.

So, AP = PB

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

Answer 2

Let Two circle's with the same centre O and AB is a chord of the larger circle toughing the smaller circle at C.


To prove : AC = BC


Construction : Join OC.


Proof : AB is a tangent to the smaller circle at the point C and OC is the radius thorough C.


Therefore, OC perpendicular AB.


The perpendicular drawn from the centre of a circle to a chord bisects the chord.


Therefore,

OC bisect AB [ Since AB is a chord of larger circle]


Hence,


AC = BC.