Determine a circle through three points

The unknowns are $A,B,C$, not the coordinates of the given points. Substituting $(3,-1)$ gives $3A-B+C=-10$; $(-2,4)$ gives $-2A+4B+C=-20$; $(6,8)$ gives $6A+8B+C=-100$. These equations are linear in $A,B,C$ even though the circle equation is quadratic in $x,y$.

The printed solution is $A=-6$, $B=-8$, $C=0$, so $x^2+y^2-6x-8y=0$. Following the source’s callout, complete the square: $(x-3)^2-9+(y-4)^2-16=0$, hence $(x-3)^2+(y-4)^2=25$. The centre is $(3,4)$ and radius $5$. Check squared distances of the three points from the centre: $25$, $25$, $25$.