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. 

Answer 1

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