Respuesta :

We need to complete the perfect square in x and y, that is,

[tex]x^2-2x=(x-1)^2-1[/tex]

and

[tex]y^2+4y=(y+2)^2-4[/tex]

Then, our given equation can be rewritten as:

[tex](x-1)^2-1+(y+2)^2-4=-1[/tex]

which is equal to

[tex]\begin{gathered} (x-1)^2+(y+2)^2-5=-1 \\ \end{gathered}[/tex]

By moving -5 to the right hand side, we have

[tex]\begin{gathered} \mleft(x-1\mright)^2+\mleft(y+2\mright)^2=-1+5 \\ \mleft(x-1\mright)^2+\mleft(y+2\mright)^2=4 \\ \mleft(x-1\mright)^2+\mleft(y+2\mright)^2=2^2 \end{gathered}[/tex]

Since the general circle equation is

[tex]\mleft(x-h\mright)^2+\mleft(y-k\mright)^2=r^2[/tex]

then, the answer is:

[tex](x-1)^2+(y+2)^2=2^2[/tex]

then, the center is (h,k)= (1, -2) and the radius is r=2