If $\alpha , \beta$ are the roots of $x^2+2x-3=0\text{ and }\gamma ,\delta$ are the roots of $y^2-y+4=0,$ then the equation of the circle having $(\alpha ,\gamma )\text{ and }(\beta ,\delta )$ as ends of diameter is 

$\alpha , \beta$ are the roots of $x^2+2x-3=0\Rightarrow x^2+2x-3=(x-\alpha )(x-\beta )$

$\gamma , \delta$ are the roots of $y^2-y+4=0\Rightarrow y^2-y+4=(y-\gamma )(y-\delta )$

Equation of the circle having $(\alpha ,\gamma )$ and $(\beta ,\delta )$ as ends of diameter is 

$\begin{array}{l}(x-\alpha )(x-\beta )+(y-\gamma )(y-\delta )=0\Rightarrow \left(x^2+2x-3\right)+\left(y^2-y+4\right)=0\\ \Rightarrow x^2+y^2+2x-y+1=0\end{array}$