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Question

If the two circles x2+y2+2gx+2fy=0 and x2+y2+2g1x+2f1y=0 touch each other then


A

ff1=gg1

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B

(fg_1 = f_1g\)

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C

f2+g2=f21+f21

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D

fg1=f1g1

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Solution

The correct option is B

(fg_1 = f_1g\)


|r1±r2|=C1C2|g2+f2±g21+f21|=(gg1)2+(ff1)2gf1=fg1


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