Prove that the tangent to a circle at any point on it is perpendicular to the radius that passes through the point of contact.
Given: AB is a tangent at the point P of a circle with center O and OP is a radius through the point P. To prove: OP and AB are perpendicular to each other i.e. OP ⊥ AB.
Construction: Any other point Q is taken on the tangent AB, O, Q are joined.
Proof: Any other point on AB except P is outside the circle; ∴ OQ intersects the circle at a point. Let R be the point of intersection.
∴ OR < OQ [∴ R is a point between O, Q]
Again, OR = OP [∴ radii of the same circle]
∴ OP < OQ
∴ The point Q is any point on AB,
∴ OP is the least of all the line segments drawn from the center O to the tangent AB.
Again, the least distance is perpendicular distance.
∴ OP ⊥ AB (proved)
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