find the radius and center of the circle whose eqation is X to the second power +(y-9) to the second power=24
thankss
2007-04-14 08:51:36 · 9 answers · asked by J j 2 in Science & Mathematics ➔ Mathematics
You meant
x^2 + (y - 9)^2 = 24
Radius: sqrt(24) = 2 * sqrt(6)
Center: (0, 9)
2007-04-14 08:56:22 · answer #1 · answered by jcastro 6 · 0⤊ 0⤋
The center of the circle is where x=0 and (y-9)=0. That's at (0,9).
The radius is simply the square root of 24, or 2*sqrt(6). An easy way to see this is by figuring out what x has to be when y=9. The equation turns into x^2=24, so x can be sqrt(24) or -sqrt(24).
Hope that helps!
2007-04-14 08:58:49 · answer #2 · answered by Bramblyspam 7 · 0⤊ 0⤋
x^2 + (y - 9)^2 = 24 the main equation is x^2+y^2=R^2
so the y-9 is Y so the center is (0,9) and the radius is 24^0.5
2007-04-14 09:08:36 · answer #3 · answered by suerena 2 · 0⤊ 0⤋
radius = √24
center at 0, +9
2007-04-14 08:57:46 · answer #4 · answered by Steve 7 · 0⤊ 0⤋
x^2 + (y-9)^2 = 24
center at (0,9) with radius sqrt(24)
eqn is (x-h)^2 + (y-k)^2 = r^2
radius r and center (h,k)
2007-04-14 08:57:03 · answer #5 · answered by Anonymous · 0⤊ 0⤋
The radius is the sq. root of 24
The center is the point (0,9)
2007-04-14 08:56:24 · answer #6 · answered by Scott H 3 · 0⤊ 0⤋
center is (0,9)
r=root 24
(x-a)^2+(y-b)^2=r^2
(a,b) is center
r is radius
2007-04-14 08:57:08 · answer #7 · answered by liuxiaojianshang 1 · 0⤊ 0⤋
(x-8)^2+(y-17)^2=13 (x^2-8^2)+(y^2-17^2)=13 (x^2+sixty 4)+(y^2+289)=13 x^2+y^2+353=13 x^2+y^2=-340 this is the suitable i'm able to do. i've got faith that's reliable and you may attempt to resolve it from here.
2016-11-23 19:54:17 · answer #8 · answered by ? 4 · 0⤊ 0⤋
r=sqrt(24)=2sqrt(6)
center=(0,9)
2007-04-14 08:56:28 · answer #9 · answered by ooorah 6 · 0⤊ 0⤋