SOLUTION: How could I find the center (h,k) and radius r of this equation: 2(x-3)^2 + 2y^2=8 ?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: How could I find the center (h,k) and radius r of this equation: 2(x-3)^2 + 2y^2=8 ?      Log On


   

Question 397311: How could I find the center (h,k) and radius r of this equation: 2(x-3)^2 + 2y^2=8 ?
Found 2 solutions by solver91311, CharlesG2:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!




is the equation of a circle centered at with radius

First, multiply both sides of your equation by



Then re-write your equation thus:



Then your answers are available by inspection.

John

My calculator said it, I believe it, that settles it
The Out Campaign: Scarlet Letter of Atheism

Answer by CharlesG2(834) About Me  (Show Source):
You can put this solution on YOUR website!
How could I find the center (h,k) and radius r of this equation: 2(x-3)^2 + 2y^2=8 ?
standard equation of a circle with center (h,k) and radius r:
(x - h)^2 + (y - k)^2 = r^2

2(x - 3)^2 + 2y^2 = 8
(x - 3)^2 + y^2 = 4
(x - 3)^2 + (y - 0)^2 = 4
center is (3,0), radius = 2