Question

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

Answer 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`

Answer 2

Given: A circle C (O, 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.

Answer 3

Given, a circle with center O and tangent at P.

To prove: OP ⊥ PQ

Constant: Extend OR to Q, at AB

Proof: We have

OP = OR   ...(Radius)

OQ = OR + RQ

Clearly OQ > OR

∴ OQ > OP

The shortest line joining a point to any point on given line is ⊥r to that line

⇒ OP ⊥ AB

or OP ⊥ PQ

Hence Proved.