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Sagot :
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
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