# What is the center and radius of the circle with equation 2(x-2)^2+2(y+5)^2=28?

Jan 25, 2016

Center $\left(x , y\right) = \left(2 , - 5\right)$
Radius: $\sqrt{14}$

#### Explanation:

$2 {\left(x - 2\right)}^{2} + 2 {\left(y + 5\right)}^{2} = 28$
$\textcolor{w h i t e}{\text{XXX}}$is equivalent to
${\left(x - 2\right)}^{2} + {\left(y + 5\right)}^{2} = 14$ (after dividing by $2$)
or
${\left(x - 2\right)}^{2} + {\left(y - \left(- 5\right)\right)}^{2} = {\left(\sqrt{14}\right)}^{2}$

Any equation of the form
$\textcolor{w h i t e}{\text{XXX}} {\left(x - a\right)}^{2} + \left(y - b\right) 2 = {r}^{2}$
is a circle with center $\left(a , b\right)$ and radius $r$

So the given equation
is a circle with center $\left(2 , - 5\right)$ and radius $\sqrt{14}$
graph{2(x-2)^2+2(y+5)^2=28 [-7.78, 10, -8.82, 0.07]}