We are going to develop the mechanism to obtain the equation of a circumference if we know that it goes through three different points.
The general equation of a circumference is $$x^2+y^2+Ax+Bx+C=0$$ and has $$3$$ parameters to be determined: $$A$$, $$B$$ and $$C$$.
Therefore, we know that if it is a system of 3 equations, we will be able to determine the three parameters.
And so, we have to use the three given points that we know, which are of the circumference, and substituting them in the general equation we get three equations with three unknowns $$A$$, $$B$$ and $$C$$.
Let's suppose that the circumference goes through points $$(0,0)$$, $$(3,1)$$ and $$(5,7)$$.
We put each of the three points in the general equation of the circumference: $$$(0,0) \Rightarrow 0^2+0^2+A \cdot 0 + B \cdot 0 + C = 0 \\ (3,1) \Rightarrow 3^2+1^2+A \cdot 3+B \cdot 1+C=0 \\ (5,7) \Rightarrow 5^2+7^2+A \cdot 5 + B \cdot 7 +C =0$$$ We must solve the following equations system for unknowns $$A$$, $$B$$ and $$C$$: $$$ \left \{{\begin{array}{r} C=0 \\ 9+1+A \cdot 3+B \cdot 1+C=0 \\ 25+49+A \cdot 5 + B \cdot 7 +C=0 \end{array}}\right.$$$
First, we replace the $$C$$ in other equations since we know it (it is zero) and we obtain a system of only two equations and two unknowns. $$$ \left\{{\begin{array}{r}10+3A+B=0 \\ 74+5A+7B=0 \end{array} }\right.$$$
We isolate $$B$$ from the first equation: $$$B-10-3A$$$ and we put it into the second equation from which we will be able to isolate and obtain $$A$$: $$$\displaystyle \begin{array}{r}74+5A+7(-10-3A)=0 \\ 74+5A-70-21A=0 \\ 16A=4 \\ A= \frac{4}{16}=\frac{1}{4}\end{array}$$$
Then we replace the value of $$A$$ obtained in the expression $$$B=-10-3A$$$ and we will get $$$\displaystyle B=-\frac{43}{4}$$$
So we know already each of the parameters that determine the circumference, therefore we can write the equation: $$$\displaystyle x^2+y^2+\frac{1}{4}x-\frac{4}{43}=0$$$