Respuesta :
Answer:
The center of the circle is at;
[tex](1,-3)[/tex]The radius of the circle is;
[tex]r=4[/tex]Explanation:
Given the equation of circle;
[tex]x^2+y^2-2x+6y-6=0[/tex]we want to re-write it in the form;
[tex](x-h)^2+(y-k)^2=r^2[/tex]where;
[tex]\begin{gathered} (h,k)\text{ is the center of the circle} \\ r\text{ is the radius} \end{gathered}[/tex]Applying Completing the square method;
[tex]\begin{gathered} x^2+y^2-2x+6y-6=0 \\ x^2-2x+1+y^2+6y+9=6+1+9 \\ (x-1)^2+(y+3)^2=16 \\ (x-1)^2+(y+3)^2=4^2 \end{gathered}[/tex]comparing the derived equation to the general form we have;
[tex]\begin{gathered} h=1 \\ k=-3 \\ r=4 \end{gathered}[/tex]Therefore;
The center of the circle is at;
[tex](1,-3)[/tex]The radius of the circle is;
[tex]r=4[/tex]