Question

Prove that a tangent surrounded by a circle of a circle is perpendicular to the radius passing through the tangent.​

Answer 1

Step-by-step explanation:

circle of a circle is perpendicular to the

Answer 2

Answer:

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. centre of the circle to a point on l (l is the tangent to the circle), the perpendicular is shortest to l.

O is the centre 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.

Theorem – The tangent at any point of a circle is perpendicular to the radius through the point of contact – Circles |

Tangent is a straight line drawn from an external point that touches a circle at exactly one point on the circumference of the circle. There can be an infinite number of tangents of a circle. These tangents follow certain properties that can be used as identities to perform mathematical computations on circles.

Here, in this article, we will learn about one of such properties i.e. the tangent at any point of a circle is perpendicular to the radius through the point of contact.

To Prove: The tangent at any point of a circle is perpendicular to the radius through the point of contact

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

Step-by-step explanation:

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