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