If alpha and beta are the roots of x^2+2x-3=0 | vedclass

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(A) For the equation $x^2+2x-3=0$,the roots are $\alpha$ and $\beta$. By Vieta's formulas,$\alpha+\beta = -2$ and $\alpha\beta = -3$.For the equation

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$x^2+y^2+2x-y-9=0$

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(A) For the equation $x^2+2x-3=0$,the roots are $\alpha$ and $\beta$. By Vieta's formulas,$\alpha+\beta = -2$ and $\alpha\beta = -3$.For the equation $y^2-y-6=0$,the roots are $\gamma$ and $\delta$. By Vieta's formulas,$\gamma+\delta = 1$ and $\gamma\delta = -6$.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.Expanding this,we get $x^2 - (x_1+x_2)x + x_1x_2 + y^2 - (y_1+y_2)y + y_1y_2 = 0$.Substituting the values,we get $x^2 - (-2)x + (-3) + y^2 - (1)y + (-6) = 0$.This simplifies to $x^2 + y^2 + 2x - y - 9 = 0$.

If $\alpha$ and $\beta$ are the roots of $x^2+2x-3=0$ and $\gamma$ and $\delta$ are the roots of $y^2-y-6=0$,then the equation of the circle having $(\alpha, \gamma)$ and $(\beta, \delta)$ as ends of a diameter is

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