Two circles of radius 4cm, and 9cm touch each other externally. A common tangent AB is drawn to these two circles, A and B lie on the two circles respectively. Then the radius of the circle which touches the given circles, and also whose tangent is line AB is(a) 1.32 cm (b) 1.44 cm (c) 1.56 cm (d) 1.68 cm

• b) 1.44 cm
  
  One way of solving is
  9 and 4 are perfect square=>radius of other circle also a perfect square=1.44cm
  
  The other way is
  L = √(4×9×4) = 12 cm
  The new circle is b/w the tangent and two circles and radius = r
  
  The length of tangent b/w r and 9cm circles = √(4×r×9) = L1
  The Length of tangent b/w r and 4cm circles = √(4×r×4) = L2
  Also (L1 + L2) = 12
  => √(4×r×9)+ √(4×r×4) = 12
  => 10√r = 12
  => r = 1.44 cm
  
  
• 11 years agoHelpfull: Yes(2) No(1)