Common Tangents
18
Pearson Malaysia Sdn Bhd Form 4 Chapter 8: Circles III
description
Common Tangents. Two circles which intersect at two points. Common tangent:. AB and CD. Q. Properties:. P. Parallel to PQ Same length that is AB = CD. Common Tangents. A. B. C. D. Two circles which intersect at two points. A. B. Q. Common tangent:. D. C. P. AB and CD. - PowerPoint PPT Presentation
Transcript of Common Tangents
Pearson Malaysia Sdn Bhd
Form 4 Chapter 8: Circles III
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Two circles which intersect at two points
P Q
A B
C D
Common tangent:
AB and CD
Properties:
• Parallel to PQ• Same length that is
AB = CD
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P Q
AB
C
D Common tangent:
AB and CD
F
Two circles which intersect at two points
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P Q
AB
C
D
F
Properties:
• Intersect at point F• AB = CD
Two cirlces which intersect at two points
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PQ
A
B
F
Properties:
Perpendicular to FQP
Common tangent:
AB
Two circles which intersect at only one point
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P Q
Common tangent:
AB and CD
Properties:
• Parallel to PQ• AB = CD
A B
C D
Two circles which intersect at only one point
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P Q
A B
C D
Common tangent:
EF
Properties:
Perpendicular to PQ
E
F
Two circles which intersect at only one point
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P
G
Q
A
B
C
D
Common tangent:
AB and CD
Properties:
• Intersect at point G• AB = CD
Two circles which intersect at only one point
Pearson Malaysia Sdn Bhd
Common tangent:
EF
Properties:
Perpendicular to PQ
P
G
Q
A
B
C
D
E
F
Two circles which intersect at only one point
Pearson Malaysia Sdn Bhd
P Q
Common tangent:
AB and CD
Properties:
• Parallel to PQ
• AB = CD
A B
C D
Two circles which do not intersect each other
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Common tangent:
EH and FG
Properties:
• Intersect at line PQ
• EH = FG
Two circles which do not intersect each other
P Q
A B
C D
E
F
G
H
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P
A
B
C
D
FQ
Common tangent:
AB and CD
Properties:
• Intersect at point F
• Same length that is AB = CD
Two circles which do not intersect each other
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Common tangent:
GH and JK
Properties:
• Intersect at the line PQ
• Same length that is GH = JK
Two circles which do not intersect each other
P
A
B
C
D
FQ
H
G
K
J
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Solving problems
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
Q
M N
HP
x (a) the length of MN,(b) the value of x,(c) the perimeter of the shaded
region.(Assume = 3.142)
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P
Q
Solving problems
M N
Solution:
Hx
(a) PQ =
= 13 cm
PT =
= 5 cm
TIn ∆PQT,
TQ2 = PQ2 – PT2
= 132 – 52
ஃ MN = 12 cm
TQ = 12 cm
9 + 4
9 – 4
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P
Q
Solving problems
M N
Solution:
Hx
= 2.4
x = 67.4°
T
(b) tan x =
=
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P
Q
Solving problems
M N
Solution:
Hx
(c) Length of arc HM
=
HQN =
= 112.6°
T
× 2 × 3.142 × 9
Length of arc HN
= × 2 × 3.142 × 4
= 10.59 cm
= 7.86 cm
Perimeter of the shaded region =
180° – 67.4°
10.59 + 7.86 + 12 = 30.45 cm
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