Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Given: Line is tangent to the (O, r) at point A.
To prove: `bar(OA) _|_ l`
Proof: Let P ∈ l , P ≠ A.
If P is in the interior of (O,r), then the line will be a secant of the circle and not a tangent.
But l is a tangent of the circle, so P is not in the interior of the circle.
P is the point in the exterior of the circle.
OP > OA ( `bar(OA)` is the radius of the circle )
Therefore each point P ∈ l except A satisfies the inequality OP > OA.
Therefore OA is the shortest distance of line l from O.
`bar(OA) _|_ l`
Given: A circle C (0, r) and a tangent l at point A
To prove: OA ⊥ l
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 segments joining the point O to a point on l, the perpendicular is shortest to l.
OA = OC (Radius of the same circle)
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.
OA ⊥ l
Hence, the tangent at any point of a circle is perpendicular to the radius through the point of contact.