Find the equation of the circle, the coordinates of the end points of one

The equation of a circle passing through the coordinates of the end points of diameters is:

$\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)=0$

Substituting, values: $\left(x_{1}, y_{1}\right)=(-2,-3) \&\left(x_{2}, y_{2}\right)=(-3,5)$

We get

$(x+2)(x+3)+(y+3)(y-5)=0$

$\Rightarrow x^{2}+3 x+2 x+6+y^{2}-5 y+3 y-15=0$

$\Rightarrow x^{2}+y^{2}+5 x-2 y-9=0$

Ans: $x^{2}+y^{2}+5 x-2 y-9=0$