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Answer :

Let AB be the tangent drawn at the point P on the circle with O.

If possible, let PQ be perpendicular to AB, not passing through O. Join OP.

\(\therefore \) Tangent at a point to a circle is perpendicular to the radius through the point,

\(\therefore \) \( AB \ ⊥ \ OP \)

\(\Rightarrow ∠ \ OPB \ = \ 90° \)

\( ∠ \ QPB \ = \ 90° \) (Construction)

\( \therefore ∠ \ QPB \ = \ ∠ \ OPB \) , which is not possible.

\( \therefore \) It contradicts our supposition.

Hence, the perpendicular at the point of contact to the tangent to a circle passes through the centre.

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- Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the center.
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