A **straight line** **intersects** a **circle** at one or two point. The **tangent to a circle** is a line that touches the circle at one point. **Secant** intersects the circle at two points. The point at which the straight line **touches** the circle is called the **point of contact** or **point of tangency**.

Some **properties of tangents** to a circle are **Infinite number of tangents** can be drawn to a circle but **only one tangent** can be drawn at any given point on a circle. From an **external point** we can draw **two tangents** of equal length. The **radius of the circle** is **perpendicular** to the tangent at its **point of contact** and the **tangents** drawn at the **extremities** of the **diameter of a circle are parallel.**

The **tangent at any point** on a circle is perpendicular to the radius drawn to the **point of contact**.

Given: A tangent AB with point of contact P.

To prove: OP ⊥ AB

Proof:

Consider point C on AB other than P.

C must lie outside the circle. (∵ A tangent can have only one point of contact with the circle)

OC > OP (∵ C lies outside the circle)

This is true for all positions of C on AB.

Thus, OP is the shortest distance between point P and line segment AB.

Hence, OP ⊥ AB.

From an **external point** we can draw only **two tangents** to a circle.

Theorem: **Tangents drawn to a circle from an external point are equal in length**.

Given: **Two tangents** AB and AC from an external point A to points B and C on a circle.

To prove: AB = AC

Construction: Join OA, OB and OC.

Proof:

In triangles OAB and OAC,

∠OBA = 90^{o} (Radius OB ⊥ Tangent AB at B)

∠OCA = 90^{o }(Radius OC ⊥ Tangent AC at C)

In triangles OBA and OCA,

∠OBA = ∠OCA = 90^{o }

OB = OC (Radii of the same circle)

OA = OA (Common side)

Thus, ΔOBA ΔOCA (**RHS congruence rule**)

Hence, AB = AC (**Corresponding sides of congruent triangles**)

The **tangents** drawn at the **extremities** of the **diameter of a circle are parallel.**