Answer:
The center has the coordinates [tex](-1.5, 2)[/tex] and the radius 2.5
Step-by-step explanation:
Let [tex](x_0,y_0)[/tex] be the center of the circel and r be the radius, then the equation of the circle is
[tex](x-x_0)^2+(y-y_0)^2=r^2[/tex]
Circle O with center [tex](x_0,y_0)[/tex] passes through the points A(0,0), B(-3,0), and C(1, 2), so
[tex]\begin{array}{l}(0-x_0)^2+(0-y_0)^2=r^2\\ \\(-3-x_0)^2+(0-y_0)^2=r^2\\ \\(1-x_0)^2+(2-y_0)^2=r^2\end{array}\Rightarrow \begin{array}{l}x_0^2+y_0^2=r^2\\ \\(3+x_0)^2+y_0^2=r^2\\ \\(1-x_0)^2+(2-y_0)^2=r^2\end{array}[/tex]
Subtract from the second equation the first one:
[tex](3+x_0)^2+y_0^2-x_0^2-y_0^2=r^2-r^2\\ \\(3+x_0)^2-x_0^2=0\\ \\(3+x_0)^2=x_0^2\\ \\3+x_0=x_0\ \text{or}\ 3+x_0=-x_0\\ \\3=0\ \text{or}\ 2x_0=-3,\ x_0=-1.5[/tex]
Substitute it into the last two equations:
[tex]\begin{array}{l}(3-1.5)^2+y_0^2=r^2\\ \\(1+1.5)^2+(2-y_0)^2=r^2\end{array}\Rightarrow \begin{array}{l}2.25+y_0^2=r^2\\ \\6.25+(2-y_0)^2=r^2\end{array}[/tex]
Subtract them:
[tex]6.25+(2-y_0)^2-2.25-y_0^2=r^2-r^2\\ \\4+4-4y_0+y_0^2-y_0^2=0\\ \\4y_0=8\\ \\y_0=2[/tex]
Substitute into the first equation:
[tex](-1.5)^2+2^2=r^2\\ \\r^2=2.25+4\\ \\r^2=6.25\\ \\r=2.5[/tex]
So, the center has the coordinates [tex](-1.5, 2)[/tex] and the radius 2.5
