# Tangents to a Circle

#### Summary*arrow_upward*

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, ?OBA ?OCA (**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.**

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#### Questions & Answers*arrow_upward*

### 1 . Prove that any four vertices of a regular pentagon are concyclic

Given ABCDE is a regular pentagon

That is AB = BC = CD = DE = AE

Recall that the sum of angles in a regular pentagon...

### 2 . Perimeter of a right triangle is equal to the sum of the diameter

ABC is a right angled triangle right angled at B.

Hence AC is diameter of the circumcircle as angle in a semi circle is a rig...

### 3 . ABC is right angle with angel B=90".

Since BC is diameter of the circle, ?BDC = 90° [Since angle in a semi circle is a right angle]

Consider ??s ABC and ADB...

### 4 . form an external point P tangent PA and PB are drawn to a circle .If CD is the tangent to the circle at a point E and PA=15cm,find the perimeter of the triangle PCD.

Sol :

Perimeter of ?PCD = 2 x AP.

=2 x 15 = 30 cm.

### 5 . Prove that ΔPRS ~ΔPTQ. If PQ = 4cm ,PT=3cm , ST= 5cm.

Sol:

PQ = 4 cm, PT = 3 cm , ST = 5 cm

(i)

PQ x PR = PT x PS

4 x PR = 3 x 8

PR = 6 cm

QR = PR...