Two circles with centres O and O’ of radii 3 cm and 4 cm, respectively, intersect at two points P and Q, such that OP and O’P are tangents to the two circles. Find the length of the common chord PQ. $(2 m a r k s)$

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Solution

$G i v e n : O P = O Q = 4 O ’ P = O ’ Q = 3 O O ’ i s t h e p e r p e n d i c u l a r b i s e c t o r o f c h o r d P Q . L e t R b e t h e p o i n t o f i n t e r s e c t i o n o f P Q a n d O O ’ . A s s u m e P R = Q R = x a n d O R = y I n O P O ’, {(O P)}^{2} + {(O' P)}^{2} = {(O O')}^{2} \Rightarrow O O ’ = \sqrt (4^{2} + 3^{2}) = \sqrt 25 = 5 O R = y \Rightarrow O R = 5 - y I n Δ O P R, {(P R)}^{2} + {(O R)}^{2} = {(O P)}^{2} \Rightarrow x^{2} + y^{2} = 4^{2} \dots \dots \dots . . (1) I n Δ O ’ P R, {(P R)}^{2} + {(O' R)}^{2} = {(O' P)}^{2} \Rightarrow x^{2} + {(5 - y)}^{2} = 9 \dots \dots \dots \dots . (2) S u b t r a c t e q u a t i o n (2) f r o m e q u a t i o n (1) y^{2} - (25 + y^{2} - 10 y) = 16 - 9 \Rightarrow y^{2} - 25 - y^{2} + 10 y = 7 . \Rightarrow 10 y = 25 + 7 \Rightarrow 10 y = 32 \Rightarrow y = 3.2 (1 m a r k) S u b s t i t u t i n g y = 3.2 i n (1), w e g e t x = \sqrt (4^{2} - {(3.2)}^{2}) = 2.4 P Q = 2 x \Rightarrow P Q = 4.8 c m (1 m a r k)$

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Two circles with centres O and O’ of radii 3 cm and 4 cm, respectively, intersect at two points P and Q, such that OP and O’P are tangents to the two circles. Find the length of the common chord PQ.(2 marks)

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