Let the line x-y+1=0 intersect the circle x^2 | vedclass

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(C) Given the line $x-y+1=0$,we have $y=x+1$. Substituting this into the circle equation $x^2+y^2+2x+2y+1=0$: $x^2+(x+1)^2+2x+2(x+1)+1=0$ $x^2+x^2+2x+

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$c$

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(C) Given the line $x-y+1=0$,we have $y=x+1$. Substituting this into the circle equation $x^2+y^2+2x+2y+1=0$: $x^2+(x+1)^2+2x+2(x+1)+1=0$ $x^2+x^2+2x+1+2x+2x+2+1=0$ $2x^2+6x+4=0 \Rightarrow x^2+3x+2=0$ $(x+1)(x+2)=0$,so $x=-1$ or $x=-2$. For $x=-1$,$y=0$. For $x=-2$,$y=-1$. Thus,the points are $A(-1, 0)$ and $B(-2, -1)$. The equation of a circle with diameter $AB$ is $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$. $(x+1)(x+2)+(y-0)(y+1)=0$ $x^2+3x+2+y^2+y=0$ $x^2+y^2+3x+y+2=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $2g=3 \Rightarrow g=3/2$,$2f=1 \Rightarrow f=1/2$,and $c=2$. Then $g+f = 3/2 + 1/2 = 2$. Since $c=2$,we have $g+f=c$.

List-$I$	List-$II$
$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y+5=0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x=1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x+8y=27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x=9$

Let the line $x-y+1=0$ intersect the circle $x^2+y^2+2x+2y+1=0$ at two points $A$ and $B$. If $AB$ is the diameter of the circle $x^2+y^2+2gx+2fy+c=0$,then $g+f=$

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