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Question
In the following fig. , AC is a transversal common tangent to tvvo circles with centres P and Q and of radii 6cm and 3cm respectively. Given that AB = 8cm, calculate PQ.
Solution
To find :- PQ
Let BC= x cm
∴ AC= 8 + x
By Pythagcras theorem BQ = `sqrt ("x"^2 + 9)`
AC2 = PQ2 -92
`=> (8 + "x")^2 = (10 + sqrt ("x"^2 + 9))^2 - 81`
`=> 64 + "x"^2 + 16"x" = 100 + "x"^2 + 9 + 20 sqrt ("x"^2 + 9)`
`=> 5 sqrt ("x"^2 + 9) = 9 + 4"x"`
Squaring both sides
⇒ 25(x2 + 9) = 81 + l6x2 + 72x
⇒ 9x2 - 72x - 144 = 0
⇒ x2 - 8x - 16 = 0
(x - 4)2 = 0
⇒ x = 4
BQ = `sqrt (4^2 + 9) = sqrt 25 = 5`
⇒ PQ = PB + BQ
⇒ PQ = 10 + 5
⇒ PQ = 15 cm
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