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(B) Since the circle $x^2+y^2+2gx+2fy+c=0$ cuts each given circle at the extremities of a diameter,the common chord must pass through the center of th

Vedclass · Accepted Answer

$-9$

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(B) Since the circle $x^2+y^2+2gx+2fy+c=0$ cuts each given circle at the extremities of a diameter,the common chord must pass through the center of the respective circle.For the circle $x^2+y^2=4$,the center is $(0,0)$. The common chord with $x^2+y^2+2gx+2fy+c=0$ is $2gx+2fy+c+4=0$. Since it passes through $(0,0)$,we get $c+4=0$,so $c=-4$.For the circle $x^2+y^2-6x-8y+10=0$,the center is $(3,4)$. The common chord is $(2g+6)x+(2f+8)y+(c-10)=0$. Substituting $(3,4)$ and $c=-4$,we get $3(2g+6)+4(2f+8)-14=0$,which simplifies to $6g+18+8f+32-14=0$,or $6g+8f+36=0$,i.e.,$3g+4f+18=0$ $(i)$.For the circle $x^2+y^2+2x-4y-2=0$,the center is $(-1,2)$. The common chord is $(2g-2)x+(2f+4)y+(c+2)=0$. Substituting $(-1,2)$ and $c=-4$,we get $-1(2g-2)+2(2f+4)-2=0$,which simplifies to $-2g+2+4f+8-2=0$,or $-2g+4f+8=0$,i.e.,$g-2f-4=0$ $(ii)$.Solving $(i)$ and $(ii)$,we get $f=-3$ and $g=-2$.Thus,$g+f+c = -2-3-4 = -9$.

If the equation of the circle which cuts each of the circles $x^2+y^2=4$,$x^2+y^2-6x-8y+10=0$,and $x^2+y^2+2x-4y-2=0$ at the extremities of a diameter of these circles is $x^2+y^2+2gx+2fy+c=0$,then $g+f+c=$

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