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: $x-y+1=0 \quad \dots(i)$ $x^2+y^2+y-1=0 \quad \dots(ii)$ From $(i)$,we have $y = x+1$. Substituting this into $(ii)$: $x^2 +

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

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(A) Given the equations: $x-y+1=0 \quad \dots(i)$ $x^2+y^2+y-1=0 \quad \dots(ii)$ From $(i)$,we have $y = x+1$. Substituting this into $(ii)$: $x^2 + (x+1)^2 + (x+1) - 1 = 0$ $x^2 + x^2 + 2x + 1 + x + 1 - 1 = 0$ $2x^2 + 3x + 1 = 0$ $(2x+1)(x+1) = 0$ So,$x = -\frac{1}{2}$ or $x = -1$. If $x = -\frac{1}{2}$,then $y = -\frac{1}{2} + 1 = \frac{1}{2}$. Point $A = (-\frac{1}{2}, \frac{1}{2})$. If $x = -1$,then $y = -1 + 1 = 0$. Point $B = (-1, 0)$. 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 $2$: $(2x+1)(x+1) + y(2y-1) = 0$ $2x^2 + 2x + x + 1 + 2y^2 - y = 0$ $2(x^2+y^2) + 3x - y + 1 = 0$.

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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