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.