A circle passes through the vertices of a triangle ABC. If the vertices are $A(-2,5),B(-2,-3),C(2,-3)$, then the canter of the circle is

$A(-2,5),B(-2,-3)$ and $(2,-3)$

$\therefore AB=8$

$BC=4$

$AC=\sqrt{16+64}=\sqrt{80}$

$\because AB^2+BC^2=AC^2$

$\Rightarrow \therefore$ Given triangle ; is right angle triangle Right angled at B.

$\therefore$ Circumcentre will be mid point of hypotenuse

$=\left(\dfrac{-2+2}{2},\dfrac{5-3}{2}\right)=(0,1)$