# How do you find the center of the radius of (x-4)^2 +y^2 =16?

Jun 2, 2016

Center: (4,0)

#### Explanation:

The circle formula is

${\left(x - {x}_{1}\right)}^{2} + {\left(y - {y}_{1}\right)}^{2} = {r}^{2}$

The center of all circles is $\left({x}_{1} , {y}_{1}\right)$, without the negative. The radius is the r. So we must square root the ${r}^{2}$ to find r, and it will be the positive square root as length must be positive.

So from the equation

${\left(x - 4\right)}^{2} + \left({y}^{2}\right) = 16$

We can know the radius is $+ \sqrt{16} = 4$

The center is ${x}_{1} , {y}_{1}$. ${x}_{1} = 4$ and ${y}_{1} = 0$.

${y}_{1}$ is 0 as there must be a y co-ordinate of the center and the only number which allows ${\left(y - {y}_{1}\right)}^{2} = {y}^{2}$ is 0