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Let the line $x-y+1=0$ intersect the circle $x^2+y^2+2 x$ $+2 y+1=0$ in two points $A$ and $B$. If $A B$ is the diameter of the circle $x^2+y^2+2 g x+2 f y+c=0$ then $g+f=$
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Verified Answer
The correct answer is:
c
$\begin{aligned} & x-y+1=0 \Rightarrow y=x+1 \\ & x^2+y^2+2 x+2 y+1=0\end{aligned}$
$$
\begin{aligned}
& \Rightarrow x^2+(x+1)^2+2 x+2(x+1)+1=0 \\
& \Rightarrow x^2+3 x+2=0 \Rightarrow x=-1,-2 \\
& \therefore y=0,-1 \\
& \therefore A(-1,0) \text { and } B(-2,-1)
\end{aligned}
$$
Equation of circle with $A B$ as diameter.
$$
\begin{aligned}
& (x+1)(x+2)+y(y+1)=0 \\
& \Rightarrow x^2+y^2+3 x+y+2=0 \\
& \therefore \quad g=\frac{3}{2}, f=\frac{1}{2}, c=2 \\
& \quad g+f=\frac{3}{2}+\frac{1}{2}=2 \\
& \therefore \quad g+f=c .
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow x^2+(x+1)^2+2 x+2(x+1)+1=0 \\
& \Rightarrow x^2+3 x+2=0 \Rightarrow x=-1,-2 \\
& \therefore y=0,-1 \\
& \therefore A(-1,0) \text { and } B(-2,-1)
\end{aligned}
$$
Equation of circle with $A B$ as diameter.
$$
\begin{aligned}
& (x+1)(x+2)+y(y+1)=0 \\
& \Rightarrow x^2+y^2+3 x+y+2=0 \\
& \therefore \quad g=\frac{3}{2}, f=\frac{1}{2}, c=2 \\
& \quad g+f=\frac{3}{2}+\frac{1}{2}=2 \\
& \therefore \quad g+f=c .
\end{aligned}
$$
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