# How do you find the center and radius for 2(x-2)^2+2(y+5)^2=28?

May 17, 2018

The equation $2 {\left(x - 2\right)}^{2} + 2 {\left(y - 5\right)}^{2} = 28$ represents a circle with center $\left(2 , 5\right)$ and radius $\sqrt{14}$.

#### Explanation:

We can write $2 {\left(x - 2\right)}^{2} + 2 {\left(y - 5\right)}^{2} = 28$

as ${\left(x - 2\right)}^{2} + {\left(y - 5\right)}^{2} = 14$

or ${\left(x - 2\right)}^{2} + {\left(y - 5\right)}^{2} = {\left(\sqrt{14}\right)}^{2}$

or $\sqrt{{\left(x - 2\right)}^{2} + {\left(y - 5\right)}^{2}} = \sqrt{14}$

this means the point $\left(x , y\right)$ on the curve moves so that its distance $\sqrt{{\left(x - 2\right)}^{2} + {\left(y - 5\right)}^{2}}$ from point $\left(2 , 5\right)$ is always $\sqrt{14}$ i.e. it forms a circle with center $\left(2 , 5\right)$ and radius $\sqrt{14}$.

Hence, the equation $2 {\left(x - 2\right)}^{2} + 2 {\left(y - 5\right)}^{2} = 28$ represents a circle with center $\left(2 , 5\right)$ and radius $\sqrt{14}$.

graph{(2(x-2)^2+2(y-5)^2-28)((x-2)^2+(y-5)^2-0.03)=0 [-8.79, 11.21, -0.6, 9.4]}