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Question

If the circle x2+y2+2gx+2fy+c=0 cuts each of the circles x2+y24=0,x2+y26x8y+10=0 and x2+y2+2x4y2=0 at the extremities of a diameter, then

A
c=4
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B
g+f=c1
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C
g2+f2c=17
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D
gf=6
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Solution

The correct options are
A c=4
B g+f=c1
C g2+f2c=17
D gf=6
Since the circle x2+y2+2gx+2fy+c=0 cuts the three given circles at the extremities of a diameter, the common chords will pass through the centre of the respective circles, so that
2gx+2fy+c+4=0 passes through (0,0) c=4(1)
Next 2gx+2fy+c+6x+8y10=0 passess through (3,4) (2g+6)3+(2f+8)414=0 3g+4f+18=0(ii) and 2gx+2fy+c2x+4y+2=0 passes through (1,2)
(2g2)(1)+(2f+4)22=0 g2f4=0(iii)
From (ii) and (iii) we get g=2 and f=3.
g+f=c1=5
g2+f2c=4+9+4=17
gf=6,
Ans: A,B,C,D

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