# How do you find the center and radius of the circle (x-2)^2+(y+1)^2=2?

May 29, 2016

The center is $\left(2 , - 1\right)$ and the radius is $\sqrt{2}$.

#### Explanation:

The circle are all the points with a fixed distance from the center. The distance is the radius.

If a point has coordinates $\left(x , y\right)$ and the center $\left({c}_{x} , {c}_{y}\right)$ the distance between the point and the center is given by the Pitagora's theorem

$\sqrt{{\left(x - {c}_{x}\right)}^{2} + {\left(y - {c}_{y}\right)}^{2}}$

and this distance has to be equal to the radius

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

In your case you do not have the square root, so you are doing the square in both sides

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

It is enough now to compare this equation with your equation to see that

${c}_{x} = 2$, ${c}_{y} = - 1$ and ${r}^{2} = 2$.

So the center has coordinates $\left(2 , - 1\right)$ and the radius is $\sqrt{2}$.