# What is the radius and center for (x-2)^{2}+y^{2}=16? A. r=16; center

What is the radius and center for $$\displaystyle{\left({x}-{2}\right)}^{{{2}}}+{y}^{{{2}}}={16}?$$
A. $$\displaystyle{r}={16}$$; center is (2,0)
B. $$\displaystyle{r}={4}$$; center is (-2,0)
C. $$\displaystyle{r}={16}$$; center is (-2,0)
D. $$\displaystyle{r}={4}$$; center is (2,0)

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Alix Ortiz
Step 1
Given equation is $$\displaystyle{\left({x}−{2}\right)}^{{{2}}}+{y}^{{{2}}}={16}$$
Step 2
given equation is a circle
standard form of an equation of a circle is $$\displaystyle{\left({x}-{h}\right)}^{{{2}}}+{\left({y}-{k}\right)}^{{{2}}}={r}^{{{2}}}$$
center $$\displaystyle={\left({h},{k}\right)}$$
radius $$\displaystyle={r}$$
lets write the given equation in standard form
$$\displaystyle{\left({x}-{2}\right)}^{{{2}}}+{\left({y}-{0}\right)}^{{{2}}}={4}^{{{2}}}$$
center $$\displaystyle={\left({2},{0}\right)}$$
radius $$\displaystyle={r}={4}$$
Thus D is a correct option.