Question

Find the center, foci, vertices, asymptotes, and radius, if necessary, of the conic sections in the equation:
\(\displaystyle{x}^{2}+{4}{x}+{y}^{2}={12}\)

Answer 1

Since you have submitted two questions, we'll answer the first question. For the second question please submit the question again and specify it.
\(\displaystyle{x}^{2}+{4}{x}+{y}^{2}={12}\) 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