Question
Wed December 05, 2012 By:

Expert Reply
Fri December 07, 2012
SInce AC and BC are tangents therefor Angle ACB=90
AC=3 & BC=4
therfor AB=(3^2+4^2)^(1/2)=5
 
 
let AE=x
therefor EB=5-x
 
now AC^2-AE^2=CB^2-BE^2
 
or, 9-x^2=16-(5-X)^2
 
Solving it x=1.8
 
therfor , CE^2=AC^2-AE^2=9-3.24=5.76
 
CE=2.4
 
length of common chord=CD=2*CE=2*2.4=4.8
 
 
 
 
 
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