Question

prove that the tangents at any point of a circle is perpendicular to the radius through point of contact​

Answer 1

Given : A circle C (0, r) and a tangent l at point A.

To prove : OA ⊥ l

Construction : Take a point B, other than A, on the tangent l. Join OB. Suppose OB meets the circle in C.

Proof: We know that, among all line segment joining the point O to a point on l, the perpendicular is shortest to l.

OA = OC (Radius of the same circle)

Now, OB = OC + BC.

∴ OB > OC

⇒ OB > OA

⇒ OA < OB

B is an arbitrary point on the tangent l. Thus, OA is shorter than any other line segment joining O to any point on l.

Here, OA ⊥ l

Answer 2

Proof: We know that, among all line segment joining the point O to a point on l, the perpendicular is shortest to l. Now, OB = OC + BC. B is an arbitrary point on the tangent l. Thus, OA is shorter than any other line segment joining O to any point on l.