Question

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

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

Answer 1

Let us consider a circle with centre O. Let AB be a tangent which touches the circle at P.

We have to prove that the line perpendicular to AB at P passes through centre O. We shall prove this by contradiction method.

Let us assume that the perpendicular to AB at P does not pass through centre O. Let it pass through another point O’. Join OP and O’P.

As perpendicular to AB at P passes through O’, therefore,

∠OPB = 90°               … (1)

O is the centre of the circle and P is the point of contact. We know the line joining the centre and the point of contact to the tangent of the circle are perpendicular to each other.

∴ ∠OPB = 90°         … (2)

Comparing equations (1) and (2), we obtain

∠OPB = ∠OPB                 … (3)

From the figure, it can be observed that,

∠OPB < ∠OPB                 … (4)

Therefore, ∠O’PB = ∠OPB is not possible. It is only possible, when the line O’P coincides with OP.

Therefore, the perpendicular to AB through P passes through centre O.

Answer 2

Let O be the centre of the circle.

A tangent PR has been drawn, touching the circle at point P.

Draw QP ⊥ RP at point P, such that Q lies on the circle.

∠OPR = 90°              ...(Radius ⊥ tangent)

Also, ∠QPR = 90°                   ...(Given)

∴ ∠OPR = ∠QPR

Now, the above case is possible only when centre O lies on the line QP.

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