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Question
In the figure, given below, AC is a transverse common tangent to two circles with centres P and Q and of radii 6 cm and 3 cm respectively.
Given that AB = 8 cm, calculate PQ.
Solution
Since AC is tangent to the circle with center P at point A.
∴ ∠PAB = 90°
Similarly, ∠QCB = 90°
In ΔPAB and ΔQCB
∠PAB = ∠OCB = 90°
∠PBA = ∠QBC ...(Vertically opposite angles)
∴ ΔPAB ∼ ΔQCB
`=>(PA)/(QC) = (PB)/(QB)` ...(i)
Also in right ΔPAB,
`PB = sqrt(PA^2 + PB^2)`
`=> PB = sqrt(6^2 + 8^2)`
= `sqrt(36 + 64)`
= `sqrt(100)`
= 10 cm ...(ii)
From (i) and (ii)
`6/3 = (10)/(QB)`
`=> QB = (3 xx 10)/6 = 5 cm`
Now,
PQ = PB + QB
= (10 + 5) cm
= 15 cm
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