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(C) The equation of the line is $x+y=1$ $\ldots(i)$ and the equation of the curve is $x^2+y^2+2hxy+gx+fy+1=0$ $\ldots(ii)$.Making Eq. $(ii)$ homogeneo

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$x+y+4=0$

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(C) The equation of the line is $x+y=1$ $\ldots(i)$ and the equation of the curve is $x^2+y^2+2hxy+gx+fy+1=0$ $\ldots(ii)$.Making Eq. $(ii)$ homogeneous by Eq. $(i)$,we get the equation of the lines joining the origin to the point of intersection of Eqs. $(i)$ and $(ii)$:$x^2+y^2+2hxy+(gx+fy)(x+y)+1(x+y)^2=0$$\Rightarrow x^2+y^2+2hxy+gx^2+gxy+fxy+fy^2+x^2+y^2+2xy=0$$\Rightarrow (2+g)x^2+(2+f)y^2+xy(g+f+2h+2)=0$ $\ldots(iii)$Lines denoted by Eq. $(iii)$ will be perpendicular to each other if the sum of the coefficients of $x^2$ and $y^2$ is zero:$(2+g)+(2+f)=0$$\Rightarrow g+f+4=0$Thus,the locus of $(g, f)$ is $x+y+4=0$.

If the pair of lines joining the origin to the points of intersection of the line $x+y=1$ with the curve $x^2+y^2+2hxy+gx+fy+1=0$ are at right angles,then the point $(g, f)$ lies on the line

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