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\(\displaystyle{x}^{2}+{4}{x}+{y}^{2}={12}\) is a circle

First write the equation in standard form of circle

\(\displaystyle{x}^{2}+{4}{x}+{y}^{2}={12}\)

\(\displaystyle{x}^{2}+{4}{x}+{4}+{y}^{2}={12}+{4}\)

\(\displaystyle{\left({x}+{2}\right)}^{2}+{y}^{2}={16}\)

Compare the circle with the standard form of circle and find h,k,r

\(\displaystyle{\left({x}+{2}\right)}^{2}+{y}^{2}={16}\)

\(\displaystyle{\left({x}+{2}\right)}^{2}+{\left({y}-{0}\right)}^{2}={16}\)

\(\displaystyle{\left({x}-{h}\right)}^{2}+{\left({y}-{k}\right)}^{2}={r}^{2}\)

\(\displaystyle{h}=-{2},{k}={0}\)

\(\displaystyle{r}^{2}={16}\)

\(\displaystyle{r}={4}\)

Circle has center and radius.

Plug h,k,r in the formula of center and radius and find them.

Center \(\displaystyle={\left({h},{k}\right)}={\left(-{2},{0}\right)}\)

Radius \(\displaystyle={r}={4}\) units

Answer: Center \(\displaystyle={\left(-{2},{0}\right)}\)

Radius \(\displaystyle={4}\) units

\(\displaystyle{x}^{2}+{4}{x}+{y}^{2}={12}\) is a circle

First write the equation in standard form of circle

\(\displaystyle{x}^{2}+{4}{x}+{y}^{2}={12}\)

\(\displaystyle{x}^{2}+{4}{x}+{4}+{y}^{2}={12}+{4}\)

\(\displaystyle{\left({x}+{2}\right)}^{2}+{y}^{2}={16}\)

Compare the circle with the standard form of circle and find h,k,r

\(\displaystyle{\left({x}+{2}\right)}^{2}+{y}^{2}={16}\)

\(\displaystyle{\left({x}+{2}\right)}^{2}+{\left({y}-{0}\right)}^{2}={16}\)

\(\displaystyle{\left({x}-{h}\right)}^{2}+{\left({y}-{k}\right)}^{2}={r}^{2}\)

\(\displaystyle{h}=-{2},{k}={0}\)

\(\displaystyle{r}^{2}={16}\)

\(\displaystyle{r}={4}\)

Circle has center and radius.

Plug h,k,r in the formula of center and radius and find them.

Center \(\displaystyle={\left({h},{k}\right)}={\left(-{2},{0}\right)}\)

Radius \(\displaystyle={r}={4}\) units

Answer: Center \(\displaystyle={\left(-{2},{0}\right)}\)

Radius \(\displaystyle={4}\) units