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.

Answer 1

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