Page 159 - Mathematics_Form_3

Circles

ˆ
TQO TPO=90= ˆ ° x + z = 9 (3)
(radius is perpendicular to the tangent) Subtracting equation (2) from

OT = OT (common side of both equation (1) gives,

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triangles) x − z = 1 (4)
Thus, QOT∆ ≡∆ POT (RHS postulate) Adding equations (3) and (4) gives,
2x = 10
Therefore, TP = TQ (Definition of ⇒ x = 5

congruent triangles). This proves that
the tangents from a common external Substituting x = 5 in equations (1)
point are equal. and (3), gives,
y = 3 and z = 4 .
Example 5.22
Therefore, x = 5 cm, y = 3 cm , and
In the following figure, AB, BC, z = 4 cm.
and CA are tangents to the circle.
Calculate the values of x, y, and z
given that ‾ AB = 8 cm, ‾ BC = 7 cm, Example 5.23

and ‾ CA = 9 cm. Find the value of ,x y , and z in the

A following diagram. If XY and XR
are tangents to the circle centred at O.
x Y

R y 15 cm
P x z
O X X
z C R 7 cm
Mathematics for Secondary Schools Solution external point) Solution , it implies that
B
y
Q

From the figure, it implies that
In OXY∆
‾ = ‾ AR = x (equal tangents from an
AP

ˆ
° (radius is perpendicular
OYX 90=
external point)
to tangent)
‾ = ‾ BQ = y (equal tangents from an
BP
y
7
OX = +
‾ = ‾ CR = z (equal tangents from an
CQ

( ) ( ) ( )
2
2
2
OY +
YX
OX
=
Thus, external point) Apply the Pythagoras' theorem
x + y = 8 (1) ⇒ (y + 7) = 2 y + 2 15 2

y + z = 7 (2)

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