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

We get:

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

$\Rightarrow x^{2}-2 x-3 x+6+y^{2}-5 y-2 y+10=0$

$\Rightarrow x^{2}+y^{2}-5 x-7 y+16=0$

Ans: $x^{2}+y^{2}-5 x-7 y+16=0$