Question

If the circle x2+y2+2gx+2fy+c=0 cuts each of the circles x2+y2âˆ’4=0,x2+y2âˆ’6xâˆ’8y+10=0 and x2+y2+2xâˆ’4yâˆ’2=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=c−1 C g2+f2−c=17 D gf=6Since 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 that2gx+2fy+c+4=0 passes through (0,0) ⇒c=−4(1) Next 2gx+2fy+c+6x+8y−10=0 passess through (3,4) ⇒(2g+6)3+(2f+8)4−14=0 ⇒3g+4f+18=0(ii) and 2gx+2fy+c−2x+4y+2=0 passes through (−1,2)⇒(2g−2)(−1)+(2f+4)2−2=0 ⇒g−2f−4=0(iii) From (ii) and (iii) we get g=−2 and f=−3.⇒g+f=c−1=−5 g2+f2−c=4+9+4=17gf=6,Ans: A,B,C,D

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