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: $\left(x_{1}, y_{1}\right)=(5,-3) \&\left(x_{2}, y_{2}\right)=(2,-4)$

We get:

$(x-5)(x-2)+(y+3)(y+4)=0$

$\Rightarrow x^{2}-2 x-5 x+10+y^{2}+3 y+4 y+12=0$

$\Rightarrow x^{2}+y^{2}-7 x+7 y+22=0$

Ans: $x^{2}+y^{2}-7 x+7 y+22=0$