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Q.

The condition that the two circlesx2+y2+2gx+c=0;x2+y2+2fy+c=0  may touch each other is

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a

1g+1f=1c

b

1g2+1f2=2c

c

1g2+1f2=1c2

d

1g2+1f2=1c

answer is D.

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Detailed Solution

C1=(−g,o),C2=(o,−f) r1=g2−c,r2=f2−c d=|r1±r2|⇒g2+f2=|g2−c±f2−c| ⇒1g2+1f2=1c
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