Find the circumference that goes through points $$a=(2,0)$$, $$b=(2,3)$$ and $$c=(1,3)$$.
See development and solution
Development:
We replace them in the general equation of the circumference $$x^2+y^2+Ax+By+C=0$$. Thereby we obtain:
$$$\left\{\begin{array}{c} 2^2+0+2A+0+C=0 \\ 2^2+3^2+2A+3B+C=0 \\ 1^2+3^2+A+3B+C=0 \end{array}\right\} \Rightarrow \left\{\begin{array}{c} 4+2A=-C \\ 13+2A+3B+C=0 \\ 10+A+3B+C=0 \end{array}\right.$$$ We solve the system and get: $$$\left\{\begin{array}{c} A=-3 \\ B=-3 \\ C=2 \end{array}\right.$$$ And we have the equation $$$x^2+y^2-3x-3y+2=0$$$
Solution:
$$x^2+y^2-3x-3y+2=0$$