Consider the equation,

\(\displaystyle{\left({x}+{2}\right)}^{2}+{\left({y}-{1}\right)}^{2}={4}\)

The above equation is also written as follows,

\(\displaystyle{\left({x}+{2}\right)}^{2}+{\left({y}-{1}\right)}^{2}={2}^{2}\)

Now compare the above equation with the standard form of the equation of the circle, that is \(\displaystyle{\left({x}-{h}\right)}^{2}+{\left({y}-{k}\right)}^{2}={r}^{2}.\)

So,

\(\displaystyle{h}=-{2},{k}={1}{\quad\text{and}\quad}{r}={2}\)

Thus, the equation \(\displaystyle{\left({x}+{2}\right)}^{2}+{\left({y}-{1}\right)}^{2}={4}\) is the equation of circle with center at (-2, 1)

and radius \(\displaystyle{r}={2}\) as shown below,

Therefore, the graph has been shifted 1 unit upward and 2 units to the left from the standard position.

\(\displaystyle{\left({x}+{2}\right)}^{2}+{\left({y}-{1}\right)}^{2}={4}\)

The above equation is also written as follows,

\(\displaystyle{\left({x}+{2}\right)}^{2}+{\left({y}-{1}\right)}^{2}={2}^{2}\)

Now compare the above equation with the standard form of the equation of the circle, that is \(\displaystyle{\left({x}-{h}\right)}^{2}+{\left({y}-{k}\right)}^{2}={r}^{2}.\)

So,

\(\displaystyle{h}=-{2},{k}={1}{\quad\text{and}\quad}{r}={2}\)

Thus, the equation \(\displaystyle{\left({x}+{2}\right)}^{2}+{\left({y}-{1}\right)}^{2}={4}\) is the equation of circle with center at (-2, 1)

and radius \(\displaystyle{r}={2}\) as shown below,

Therefore, the graph has been shifted 1 unit upward and 2 units to the left from the standard position.