# How do you find the center and radius of the circle given (x-3)^2+y^2=16?

Oct 20, 2016

$\left(3 , 0\right)$

Radius$= 4$

#### Explanation:

Given -

${\left(x - 3\right)}^{2} + {y}^{2} = 16$

The equation of the circle with center $\left(h , k\right)$ is -

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

In our case to find the center, rewrite the equation like this-

${\left(x - 3\right)}^{2} + {\left(y - 0\right)}^{2} = 16$

Then the center is -

$\left(3 , 0\right)$

The radius is $= {\sqrt{a}}^{2} = \sqrt{16} = 4$