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(C) Let the equation of the circle be $x^2+y^2+2gx'+2fy'+c=0$. Since the circle passes through $(-1,0)$,$(-1,1)$,and $(1,1)$,we substitute these point

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$2$

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(C) Let the equation of the circle be $x^2+y^2+2gx'+2fy'+c=0$. Since the circle passes through $(-1,0)$,$(-1,1)$,and $(1,1)$,we substitute these points into the equation: $1$) For $(-1,0)$: $(-1)^2+0^2+2g(-1)+2f(0)+c=0 \implies 1-2g+c=0 \implies c=2g-1$. $2$) For $(-1,1)$: $(-1)^2+1^2+2g(-1)+2f(1)+c=0 \implies 2-2g+2f+c=0$. Substituting $c=2g-1$ into this: $2-2g+2f+2g-1=0 \implies 2f+1=0 \implies f=-1/2$. $3$) For $(1,1)$: $1^2+1^2+2g(1)+2f(1)+c=0 \implies 2+2g+2f+c=0$. Substituting $f=-1/2$ and $c=2g-1$: $2+2g+2(-1/2)+2g-1=0 \implies 2+2g-1+2g-1=0 \implies 4g=0 \implies g=0$. Then $c=2(0)-1=-1$. The equation is $x^2+y^2+0x+2(-1/2)y-1=0 \implies x^2+y^2-y-1=0$. To match the form $ax^2+ay^2+2gx+2fy-2=0$,we multiply the entire equation by $2$: $2x^2+2y^2-2y-2=0$. Comparing $2x^2+2y^2+0x-2y-2=0$ with $ax^2+ay^2+2gx+2fy-2=0$,we get $a=2$.

If the equation of the circle passing through the points $(-1,0), (-1,1), (1,1)$ is $ax^2+ay^2+2gx+2fy-2=0$,then $a=$

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