Page 148 - Mathematics_Form_3

Circles

line meets the circle at P and ‾ CP is
C E D 2 cm, find the radius of the circle.

Solution
13 cm Consider the following figure.
5 cm
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O

O
Solution r
8 cm 8 cm
Consider a right-angled triangle ODE. A B
C
Using Pythagoras’ theorem, it implies
2 cm
that
2 2 2
( ‾ OE ) + ( ‾ ED ) = ( ‾ OD ) P
2
2
2
⇒ (5 cm) + ( ‾ ED ) = (13 cm)
2 Given AB 16cm= , CP = 2cm

⇒ ( ‾ ED ) = (169 − 25 ) c m Let r be the radius of the circle. It
2
= 144 cm implies that,
2
_
⇒ ‾ ED = √ 144 c m 2 AC = BC = 16 cm 8cm= (OP
= 12 cm 2
Since the perpendicular bisector ‾ OE perpendicular bisector of a chord)
divides the chord CD into equal parts, But, OP = OB r= (radius of the
then, circle)
‾ = ‾ ED ⇒ OC = r − 2
CE
Thus, Apply the Pythagoras’ theorem using
( ) ( ) ( )
‾ = ‾ CE + ‾ ED 2 2 2
CD
∆ OBC , OB = OC + BC
= 12 cm + 12 cm
Thus,
= 24 cm
r = 2 (r − 2) + 2 8 2 Mathematics for Secondary Schools
Therefore, the length of the chord is 2 2
4 64
24 cm. ⇒ r = r − 4r ++
⇒ 4r = 68

r
⇒= 17
Example 5.18
Therefore, the radius of the circle is
A chord AB, 16 cm long is bisected 17 cm.
by a line from the centre O at C. If the

Student\s Book Form Three 141

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MATHEMATIC F3 SB.indd 141