Transcript chapter8 cicle comtan - MATHEMATICSSECONDARYSCHOOL

Form 4 Chapter 8: Circles III
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Two circles which intersect at two points
A
B
Common tangent:
AB and CD
P
Q
C
D
Properties:
• Parallel to PQ
• Same length that is
AB = CD
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Two circles which intersect at two points
A
B
F
Q
P
D
Common tangent:
C
AB and CD
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Two cirlces which intersect at two points
A
B
F
Q
P
D
Properties:
C
• Intersect at point F
• AB = CD
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Two circles which intersect at only one point
B
Common tangent:
AB
F
Q
P
Properties:
A
Perpendicular to FQP
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Two circles which intersect at only one point
A
B
Common tangent:
AB and CD
P
Q
Properties:
C
D
• Parallel to PQ
• AB = CD
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Two circles which intersect at only one point
A
E
B
Common tangent:
EF
P
Q
Properties:
C
F
D
Perpendicular to
PQ
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Two circles which intersect at only one point
A
Common tangent:
B
G
AB and CD
Q
P
D
Properties:
C
• Intersect at point G
• AB = CD
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Two circles which intersect at only one point
A
E
Common tangent:
B
G
EF
Q
P
F
D
Properties:
C
Perpendicular to
PQ
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Two circles which do not intersect each other
A
Common tangent:
B
AB and CD
P
C
Q
D
Properties:
• Parallel to PQ
• AB = CD
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Two circles which do not intersect each other
A
Common tangent:
B
E
G
EH and FG
P
Q
F
C
H
D
Properties:
• Intersect at line PQ
• EH = FG
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Two circles which do not intersect each other
A
Common tangent:
B
F
Q
P
D
C
AB and CD
Properties:
• Intersect at point F
• Same length that
is AB = CD
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Two circles which do not intersect each other
A
Common tangent:
B
G
GH and JK
J
F
Q
P
H
K
C
D
Properties:
• Intersect at the
line PQ
• Same length that
is GH = JK
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Solving problems
P
x
M
H
Q
N
In the diagram, P and Q are the
centres of two circles with radii
9 cm and 4 cm respectively. MN
is a common tangent to the
circles. Calculate
(a) the length of MN,
(b) the value of x,
(c) the perimeter of the shaded
region.
(Assume  = 3.142)
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Solving problems
(a)
Solution:
PQ = 9 + 4
= 13 cm
PT = 9 – 4
= 5 cm
P
T
M
x
In ∆PQT,
H
Q
N
TQ2 = PQ2 – PT2
= 132 – 52
TQ = 12 cm
ஃ MN = 12 cm
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Solving problems
(b)
Solution:
tan x =
=
= 2.4
x = 67.4°
P
T
M
x
H
Q
N
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Solving problems
(c)
Solution:
Length of arc HM
=
P
T
x
H
Q
× 2 × 3.142 × 9
= 10.59 cm
HQN = 180° – 67.4°
= 112.6°
Length of arc HN
=
× 2 × 3.142 × 4
= 7.86 cm
M
N
Perimeter of the shaded region = 10.59 + 7.86 + 12 = 30.45 cm
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