Question

Prove that tangent drawn at any point of a circle is perpendicular to the radius through the point of contact.

Answer 1

Given, a tangent AB at point P of the circle with center O.

To prove: $OP\perp AB$.

Construction: Join OQ where Q is a point (other than P) on AB.

Proof: Since Q is a point on the tangent AB (other than P).

$\therefore$ Q lies outside the circle.

Let OQ intersect the circle at R.

$\Rightarrow OR<OQ$.

But $OP=OR$. (Radii of the circle)



$\therefore OP<OQ$.

Thus, OP is the shortest distance than any other line segment joining O to any point of AB.

But, we know that the shortest distance between a point and a line is the perpendicular distance.

$\therefore OP\perp AB$

Hence Proved.