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Pearson Malaysia Sdn Bhd Form 4 Chapter 8: Circles III

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Pearson Malaysia Sdn Bhd Two circles which intersect at two points PQ A B C D Common tangent: AB and CD Properties: Parallel to PQ Same length that is AB = CD

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Pearson Malaysia Sdn Bhd P Q A B C D Common tangent: AB and CD F Two circles which intersect at two points

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Pearson Malaysia Sdn Bhd P Q A B C D F Properties: Intersect at point F AB = CD Two cirlces which intersect at two points

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Pearson Malaysia Sdn Bhd P Q A B F Properties: Perpendicular to FQP Common tangent: AB Two circles which intersect at only one point

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Pearson Malaysia Sdn Bhd PQ 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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Pearson Malaysia Sdn Bhd PQ 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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Pearson Malaysia Sdn Bhd 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

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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

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Pearson Malaysia Sdn Bhd P Q Common tangent: AB and CD Properties: Parallel to PQ AB = CD AB C D Two circles which do not intersect each other

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Pearson Malaysia Sdn Bhd Common tangent: EH and FG Properties: Intersect at line PQ EH = FG Two circles which do not intersect each other PQ AB C D E F G H

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Pearson Malaysia Sdn Bhd P A B C D F Q 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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Pearson Malaysia Sdn Bhd 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 F Q H G K J

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Pearson Malaysia Sdn Bhd 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 MN H P x (a) the length of MN, (b) the value of x,x, (c) the perimeter of the shaded region. (Assume = 3.142)

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Pearson Malaysia Sdn Bhd P Q Solving problems MN Solution: H x (a) PQ = = 13 cm PT = = 5 cm T In ∆ PQT, TQ 2 = PQ 2 – PT 2 = 13 2 – 5252 ஃ MN = 12 cm TQ = 12 cm – 4

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Pearson Malaysia Sdn Bhd P Q Solving problems MN Solution: H x = 2.4 x = 67.4° T (b) tan x = =

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Pearson Malaysia Sdn Bhd P Q Solving problems MN Solution: H x (c) Length of arc HM = HQN = = 112.6° T × 2 × × 9 Length of arc HN = × 2 × × 4 = cm = 7.86 cm Perimeter of the shaded region = 180° – 67.4° = cm

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Pearson Malaysia Sdn Bhd

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