We need to complete the perfect square in x and y, that is,
![x^2-2x=(x-1)^2-1](https://img.qammunity.org/2023/formulas/mathematics/high-school/byoqruy2mb2bp7dqnng0o6cqixhh3qt9mw.png)
and
![y^2+4y=(y+2)^2-4](https://img.qammunity.org/2023/formulas/mathematics/high-school/7herde5viabmqo7htvdjpli90r8pwtuql1.png)
Then, our given equation can be rewritten as:
![(x-1)^2-1+(y+2)^2-4=-1](https://img.qammunity.org/2023/formulas/mathematics/high-school/ackimupeddlxem06d4jjt5npulk05yozzz.png)
which is equal to
![\begin{gathered} (x-1)^2+(y+2)^2-5=-1 \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/o48kmwbbvzal30vnd85rxrtixpopoktg89.png)
By moving -5 to the right hand side, we have
![\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}](https://img.qammunity.org/2023/formulas/mathematics/high-school/lme3xda9xrzyxj2kqkhjdjr7yvel43tb6q.png)
Since the general circle equation is
![\mleft(x-h\mright)^2+\mleft(y-k\mright)^2=r^2](https://img.qammunity.org/2023/formulas/mathematics/college/wu0rxsedw06g5ye1b7u7qnxfn8hwcjr87d.png)
then, the answer is:
![(x-1)^2+(y+2)^2=2^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/ax2pe5wwjmh188tw745dxknulyhkk3tlkx.png)
then, the center is (h,k)= (1, -2) and the radius is r=2