The equation of the circle passing through the points (1,1), (2,-1), (3,2) is 

Let A(1, 1) B(2, -1)& C(3,2). Slope of $\stackrel{\_\_\_}{AB}$ is $\frac{-1-1}{2-1}=-2$ mid point of $\stackrel{\_\_\_}{AB}$ is $\left(\frac{3}{2},0\right)$

Equation to the perpendicular bisect or of $\stackrel{\_\_\_}{AB}$ is $y=\frac{1}{2}\left(x-\frac{3}{2}\right)\Rightarrow 2x-4y-3=0$

The centre of  $x^2+y^2-5x-y+4=0$ is $C\left(5/2, 1/2\right),$ C lies on 2x-4y-3=0

$\therefore$ Required equation is $x^2+y^2-5x-y+4=0$.

This method will help to check the verification.

 OR

The procedure to find the equation of the circle passing through three points is, substitute all three points in the standard form of equation, and solve those three equations for unknowns

finally write the equation of circle passing through the given three points