If x-y+1=0 meets the circle x^2+y^2+y-1=0 at | vedclass

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(A) Given the equations: $x-y+1=0 \implies y=x+1$ $(i)$ $x^2+y^2+y-1=0$ (ii) Substituting $(i)$ into (ii): $x^2+(x+1)^2+(x+1)-1=0$ $x^2+x^2+2x+1+x=0$

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$2(x^2+y^2)+3x-y+1=0$

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(A) Given the equations: $x-y+1=0 \implies y=x+1$ $(i)$ $x^2+y^2+y-1=0$ (ii) Substituting $(i)$ into (ii): $x^2+(x+1)^2+(x+1)-1=0$ $x^2+x^2+2x+1+x=0$ $2x^2+3x+1=0$ $(2x+1)(x+1)=0$ So,$x=-\frac{1}{2}$ or $x=-1$. For $x=-\frac{1}{2}$,$y=-\frac{1}{2}+1=\frac{1}{2}$. Point $A$ is $(-\frac{1}{2}, \frac{1}{2})$. For $x=-1$,$y=-1+1=0$. Point $B$ is $(-1, 0)$. The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$. $(x+\frac{1}{2})(x+1)+(y-\frac{1}{2})(y-0)=0$ $(x+\frac{1}{2})(x+1)+y(y-\frac{1}{2})=0$ Multiply by $2$: $(2x+1)(x+1)+2y(y-\frac{1}{2})=0$ $2x^2+2x+x+1+2y^2-y=0$ $2(x^2+y^2)+3x-y+1=0$.

If $x-y+1=0$ meets the circle $x^2+y^2+y-1=0$ at $A$ and $B$,then the equation of the circle with $AB$ as diameter is

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