Consider a triangle Delta whose two sides lie | vedclass

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(B) Let the vertices of the triangle be $A, B, C$. One vertex $A$ is the intersection of the $x$-axis $(y=0)$ and the line $x+y+1=0$,which gives $A(-1

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$x^2+y^2+x+3y=0$

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(B) Let the vertices of the triangle be $A, B, C$. One vertex $A$ is the intersection of the $x$-axis $(y=0)$ and the line $x+y+1=0$,which gives $A(-1,0)$.Let vertex $B$ lie on the line $x+y+1=0$,so $B(\alpha, -\alpha-1)$.The altitude from $B$ to $AC$ (which lies on the $x$-axis) is a vertical line passing through $H(1,1)$,so its equation is $x=1$. Since $B$ lies on this line,$\alpha=1$,thus $B(1,-2)$.Let vertex $C$ lie on the $x$-axis,so $C(\beta, 0)$.The altitude from $A(-1,0)$ to $BC$ passes through $H(1,1)$. The slope of $AH$ is $m_{AH} = \frac{1-0}{1-(-1)} = \frac{1}{2}$.The slope of $BC$ is $m_{BC} = \frac{-2-0}{1-\beta} = \frac{2}{\beta-1}$.Since $AH \perp BC$,$m_{AH} \cdot m_{BC} = -1$ $\Rightarrow \frac{1}{2} \cdot \frac{2}{\beta-1} = -1$ $\Rightarrow \beta-1 = -1$ $\Rightarrow \beta=0$. Thus $C(0,0)$.The vertices are $A(-1,0)$,$B(1,-2)$,and $C(0,0)$.The circumcircle equation is $x^2+y^2+gx+fy+c=0$. Since it passes through $(0,0)$,$c=0$.Passing through $(-1,0): 1-g=0 \Rightarrow g=1$.Passing through $(1,-2): 1+4+g-2f=0$ $\Rightarrow 5+1-2f=0$ $\Rightarrow 2f=6$ $\Rightarrow f=3$.The equation is $x^2+y^2+x+3y=0$.

Consider a triangle $\Delta$ whose two sides lie on the $x$-axis and the line $x+y+1=0$. If the orthocenter of $\Delta$ is $(1,1)$,then the equation of the circle passing through the vertices of the triangle $\Delta$ is

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