SOLUTION: 15. What are the center and radius of the circle whose equation is x^2 + (y-3)^2 = 144? (x-h)^2 + (y-k)^2 = r^2. (x-0)^2+(y-3)^2=12^2 (0,-3) r=12 I know this is incorre

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 15. What are the center and radius of the circle whose equation is x^2 + (y-3)^2 = 144? (x-h)^2 + (y-k)^2 = r^2. (x-0)^2+(y-3)^2=12^2 (0,-3) r=12 I know this is incorre      Log On


   

Question 162556: 15. What are the center and radius of the circle whose equation is x^2 + (y-3)^2 = 144?
(x-h)^2 + (y-k)^2 = r^2.
(x-0)^2+(y-3)^2=12^2
(0,-3)
r=12
I know this is incorrect, but I cant find where my mistake is at. If someone could point it out, I would realy apperciate it. Thank you so much.
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
I believe the (y-3)^2 should be (y+3) because (y-k)=y-(-3)=y+3.