Question

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

Answer 1

The equation `2(x-2)^2+2(y-5)^2=28` represents a circle with center `(2,5)` and radius `sqrt14`.

Explanation:

We can write `2(x-2)^2+2(y-5)^2=28`

as `(x-2)^2+(y-5)^2=14`

or `(x-2)^2+(y-5)^2=(sqrt14)^2`

or `sqrt((x-2)^2+(y-5)^2)=sqrt14`

this means the point `(x,y)` on the curve moves so that its distance `sqrt((x-2)^2+(y-5)^2)` from point `(2,5)` is always `sqrt14` i.e. it forms a circle with center `(2,5)` and radius `sqrt14`.

Hence, the equation `2(x-2)^2+2(y-5)^2=28` represents a circle with center `(2,5)` and radius `sqrt14`.

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]}