A Tangent to a circle is a line which meets the circle at one point only.

**Theorem 1:** The tangent to a circle at any point is perpendicular to the radius.

**Given:** TA is a tangent at T to a circle with centre O

**To prove:** ∠OTA = 90°

### Proof:

Let C be any other point on TA

Then |OC| > |OT|

This implies that |OT| is the shortest distance from O to |CT| since all other points on the line TCA lie outside the circle. But the shortest distance between a point and a line is the perpendicular distance. Hence |OT| is perpendicular to TA and ∠OTA = 90°

**Theorem 2: **The tangents to a circle from external points are equal.

**Given: **P is any point outside a circle center O. The tangents from P touch the circle at X and Y respectively.

**To prove: ** |XP| = |YP|

**Construction:** join OX, OY and OP

### Proof:

|OX| = |OY| **(radii)**

∠OXP= ∠OYP= 90° **(tangent perpendicular to radius)**

OP is common to the two triangles

Therefore *ΔOXP ≡ ΔOYP*

*ΔOXP = ΔOYP*

Hence, |XP| = |YP|

### Example

The diagram below shows a circle, centre O. Points A, B, C and D lie on the circumference of the circle. EDF is a tangent to the circle.*∠ABC* = 82° and angle *∠ODC* = 57°

(a.) Work out the value of *x*.

(b.) Work out the value of *y*.

**Solution:**

**Solution:**

**Consider △DOC**

⇒ \( \scriptsize \overline{OD} = \overline{OC} \)**(radii of the same circle)**

Thus, *ΔDOC* is an isosceles triangle.

∠ODC = ∠OCD = 57° **(base ∠s of isosceles △DOC)**

∠ODC + ∠OCD + x = 180° **(sum of ∠s in △DOC)**

∴ 57° + 57° + x = 180

x = 180 – 114

x = 66°

Let ∠ADO = z

**Consider Cyclic quadrilateral △ABCD**

∠ABC = 82°

∠CDA = ∠ADO + ∠ODC

∠CDA = z + 57°

but ∠CDA + ∠ABC = 180°** (interior opposite ∠s of a cyclic quadrilateral are supplementary)**

∴ z + 57° + 82° = 180°

z = 180° – 139°

z = 41°

∠ODF = 90° **(tangent perpendicular to radius)**

y + ∠ADO + ∠ODF = 180° **(angles on a straight line)**

y + z + 90° = 180°

y + 41° + 90° = 180°

y + 131° = 180°

y = 180°- 131°

y = 49°

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