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.
