Acircle contains the ordered pairs (-1,2), (0,1), and (-2,-1). find the equation of the circle.

We know that

the general form of a circle is
x²+y²+Dx+Ey+F=0

A circle contains the ordered pairs (-1,2), (0,1), and (-2,-1)

for the point (-1,2)
x=-1
y=2
x²+y²+Dx+Ey+F=0--------> (-1)²+(2)²+D*(-1)+E*(2)+F=0---------> 5-D+2*E+F=0
D=5+2E+F---------> equation 1

for the point (0,1)
x=0
y=1
x²+y²+Dx+Ey+F=0--------> (0)²+(1)²+D*(0)+E*(1)+F=0---------> 1+E+F=0
1+E+F=0----------> equation 2

for the point (-2,-1)
x=-2
y=-1
x²+y²+Dx+Ey+F=0--------> (-2)²+(-1)²+D*(-2)+E*(-1)+F=0---------> 5-2*D-E+F=0
5-2*D-E+F=0--------> equation 3

I substitute 1 in 3
D=5+2E+F
5-2*D-E+F=0-------> 5-2*[5+2E+F]-E+F=0

5-10-4E-2F-E+F=0--------> -5-5E-F=0---------> equation 4

resolve 2 and 4
1+E+F=0
-5-5E-F=0

using a graph tool
E=-1
F=0
D=5+2E+F------> 5+2*(-1)+0------> D=3

the equation of a circle is
x²+y²+Dx+Ey+F=0--------> x²+y²+3x-y=0--------> (x²+3x)+(y²-y)=0
(x²+3x)+(y²-y)=0--------> (x+3/2)²+(y-1/2)²=(9/4)+(1/4)

(x+3/2)²+(y-1/2)²=(10/4)

the center is (-1.5,0.5)
radius r=√(10/4)

the answer is
(x+3/2)²+(y-1/2)²=(10/4)

see the attached figure