Topic Content:
- Tangents to a Circle
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 centre 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 8.3.1:
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:

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°