# How do you find the center and radius for  (x-2)^2+ (y+7)^2 =40?

Aug 14, 2016

centre = (2 ,-7)
radius = $2 \sqrt{10}$

#### Explanation:

The standard form of the equation of a circle is.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where (a ,b) are the coordinates of the centre and r, the radius.

The equation here ${\left(x - 2\right)}^{2} + {\left(y + 7\right)}^{2} = 40 \text{ is in this form}$

and by comparison : a = 2 , b = -7 and ${r}^{2} = 40$

$\Rightarrow \text{ centre" =(2,-7)" and } r = \sqrt{40} = 2 \sqrt{10}$