# What are the center and radius of the circle with equation (x-3)^2 + (y+1)^2=16?

Dec 6, 2015

Center: $\left(3 , - 1\right)$.

Radius: $4$

#### Explanation:

This equation is already written in the "center-radius" form: in fact, if the center $\left({x}_{0} , {y}_{0}\right)$ and the radius $r$ are known, then the equation of the cirle is

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

So, in your case, the center is $\left(3 , - 1\right)$, and the radius is $\sqrt{16} = 4$