Prove that the perpendicular at the point of contact of the tangent to a circle passes through the centre.

Let there be a circle with centre O and BR as tangent with the point of contact as B

Let AB be the line perpendicular to BR

⇒ ∠ABR = 90° …(i)

As OB is the radius of the circle and we know that radius is perpendicular to the tangent at the point of contact

OB is perpendicular to BR

⇒ ∠OBR = 90° …(ii)

Equation (i) and (ii) implies that

⇒ ∠ABR = ∠OBR

This is only possible iff A and O lie on the same line or A and O are the same points

Case 1: Suppose A and O are on the same line

If A and O are on the same line, then the perpendicular AB to tangent BR has passed through the centre

Case 2: suppose A and O are the same points

As O itself is the centre of the circle, and A and O are the same points hence the perpendicular to the tangent at the point of contact passes through the circle

In any scenario, the line has to pass through the centre.

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