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

#### Solution 1

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.

Also P≠A

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`

#### Solution 2

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 segments 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*

Hence, the tangent at any point of a circle is perpendicular to the radius through the point of contact.