Question

the tangent at the POINT of a CIRCLE is perpendicular TO the radius through the POINT of the contact ​

Answer 1

$\huge{\underline{\sf{Answer:-}}}}$

The same will be the case for all other points on the tangent (l). So OA is shorter than any other line segment joining O to any point on l. Hence, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Answer 2

Answer:

To proof :- The tangent at the point of a circle is perpendicular to the radius through the point of the contact .

Let there be a circle C (0, r) and a tangent l at point A.

Construction :-

Step 1: Take any point B online l, other than A.

Step 2: Join OB.

Step 3: Let us say that OB meets the circle in C.

Proof :-

From prior knowledge, We know that, among all line segments joining the point O i.e. center of the circle to a point on l (l is the tangent to the circle), the perpendicular is shortest to l.

O is the center of the circle and the radius of the circle will be of fixed length hence we can say that:

OC = OA (radius)

Also OB = OC + BC.

So OC < OB.

⇒ OA < OB (since OA = OC).

The same will be the case for all other points on the tangent (l).

So OA is shorter than any other line segment joining O to any point on l.

Hence, OA ⊥ l

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