Answer :
We know that one form of the equation of the circle is:
[tex]x^2+y^2+ax+by+c=0[/tex]then, if we know that 3 points are on the circle, we can use them to find a system of equation. First, for point (2,1) we have:
[tex]\begin{gathered} (x,y)=(2,1) \\ \Rightarrow(2)^2+(1)^2+2a+b+c=0 \\ \Rightarrow2a+2+c=-5 \end{gathered}[/tex]for point (-5,-6) we have:
[tex]\begin{gathered} (-5)^2+(-6)^2-5a-6b+c=0 \\ \Rightarrow-5a-6b+c=-61 \end{gathered}[/tex]finally, for point (9,-6), the equation is:
[tex]\begin{gathered} (9)^2+(-6)^2+9a-6b+c=0 \\ \Rightarrow9a-6b+c=-117 \end{gathered}[/tex]then, we have the following system of equations:
[tex]\begin{gathered} 2a+b+c=-5 \\ -5a-6b+c=-61 \\ 9a-6b+c=-117 \end{gathered}[/tex]which has the solution:
[tex]\begin{gathered} a=-4 \\ b=12 \\ c=-9 \end{gathered}[/tex]therefore, the equation of the circle that passes through the points (2,1), (-5,-6), and (9,-6) is:
[tex]x^2+y^2-4x+12y-9=0[/tex]