Q : 1    Two circles of radii \small 5\hspace{1mm}cm and \small 3\hspace{1mm}cm intersect at two points and the distance between their  centres is \small 4\hspace{1mm}cm. Find the length of the common chord.
 

Answers (1)

Given: Two circles of radii \small 5\hspace{1mm}cm and \small 3\hspace{1mm}cm intersect at two points and the distance between their centres is \small 4\hspace{1mm}cm.

To find the length of the common chord.

Construction: Join OP and draw OM\perp AB\, \, and \, \, \, ON\perp CD.

Proof: AB is a chord of circle C(P,3)  and PM is the bisector of chord AB.

           \therefore PM\perp AB

             \angle PMA=90 \degree

Let, PM = x , so QM=4-x

In \triangleAPM, using Pythagoras theorem

          AM^2=AP^2-PM^2...........................1

Also, 

In \triangleAQM, using Pythagoras theorem

          AM^2=AQ^2-MQ^2...........................2

From 1 and 2, we get

        AP^2-PM^2=AQ^2-MQ^2

\Rightarrow 3^2-x^2=5^2-(4-x)^2

\Rightarrow 9-x^2=25-16-x^2+8x

\Rightarrow 9=9+8x

\Rightarrow 8x=0

\Rightarrow x=0

Put,x=0 in equation 1

          AM^2=3^2-0^2=9

       \Rightarrow AM=3

\Rightarrow AB=2AM=6

 

 

 

 

 

 

 

 

 

 

 

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