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Question

Prove that the length of the common chord of the two circles whose equations are (xa)2+(yb)2=c2 and (xb)2+(ya)2=c2 is 4c22(ab)2.
Hence find the condition that the two circles may touch.

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Solution

As the radius of the given circles are equal so the common chord will bisect the line joining their centres

OO=(ab)2+(ba)2OO=(ab)2+(ab)2OO=2|ab|

Now OO=2AB

OB=2|ab|2=|ab|2

In AOB , OA2=OB2+AB2

c2=(|ab|2)2+AB2AB2=c2(ab)22AB2=2c2(ab)22AB=2c2(ab)22=4c22(ab)22

Length of common chord =AC=2AB

AC=2×4c22(ab)22=4c22(ab)2

Hence proved.


702096_641414_ans_47642ad8065348108ec1e513ce1e0ffa.png

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