Question

Prove that the line segment joining the points of contact of two parallel tangents of a circle passes through its centre.

Answer 1

Given : l and m are tangents such that l || m
poin of contact of l and are are A and B

To prove : AB passes through the centre
i.e AB is a diametre.

Construction : Take a point x on the tangent l
and a point y on the tangent m

Proof : <XAO = 90° - [1]
<YBO = 90° - [2]
since we know that angle between point of
contact and radius is 90°

Adding eq. [1] and [2]
90° + 90° = 180°
therefore we know that AB is a straight line as
diametre which passes through the centre.


Thank you!!

Answer 2

Given : CD & EF are two parallel tangents at a point A and B of a circle with centre O

To Prove : APB is a diameter of the circle

Construction : Join OA and OB . Draw OG II CD

Proof - OG II CD and AO cuts them

therefore ,

∠CAO + ∠GOA = 180°

90 + ∠GOA = 180° [ OA perpendicular to CD ]

∠GOA = 90°

Similarly , ∠GOB = 90 °

=> ∠GOA + ∠GOB = 180°

=> AOB is straight line which is passing through the circle with centre O

Hence , AOB is a diameter of circle having centre O .

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