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

##### 1 Answer

May 17, 2018

#### Answer:

The equation

#### Explanation:

We can write

as

or

or

this means the point

Hence, the equation

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]}