Question

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

Answer 1

Center `(x,y)=(2,-5)`
Radius: `sqrt(14)`

Explanation:

`2(x-2)^2+2(y+5)^2=28`
`color(white)("XXX")`is equivalent to
`(x-2)^2+(y+5)^2=14` (after dividing by `2`)
or
`(x-2)^2+(y-(-5))^2=(sqrt(14))^2`

Any equation of the form
`color(white)("XXX")(x-a)^2+(y-b)2 = r^2`
is a circle with center `(a,b)` and radius `r`

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