How do you find the center and radius of (x+2)^2+(y+4)^2=256?

Apr 18, 2016

centre=(-2 , -4) , r = 16

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 coords of centre and r , the radius.

${\left(x + 2\right)}^{2} + {\left(y + 4\right)}^{2} = 256 \text{ is in this form }$

and by comparison: a = -2 , b = -4 and $r = \sqrt{256} = 16$

hence : centre = (-2 ,-4) and radius = 16