Lesson 3, Topic 3
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# Tangents to a Circle

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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:

∠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°

∠ABC = 82°

∠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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