its equation is x2 + (y - 4)2 = 4 so do I assume it is (x-0)^2+(y-4)^2?
center is (0,4) and the R=4?
2007-08-26 05:28:23 · 5 answers · asked by m_carl 1 in Science & Mathematics ➔ Mathematics
Almost. Yes, the center is (0,4), but the radius is 2, the square root of 4.
2007-08-26 05:31:56 · answer #1 · answered by Marley K 7 · 1⤊ 0⤋
Hey there!
Think the equation was in the form (x-h)^2+(y-k)^2=r^2, where r is the radius of the circle and (h,k) is the center.
So, the equation x^2+(y-4)^2=4, can be rewritten as (x-0)^2+(y-4)^2=2^2.
By using the standard form, the center is (0,4), with radius 2, not 4.
Hope it helps!
2007-08-26 05:59:34 · answer #2 · answered by ? 6 · 0⤊ 0⤋
Center is (0,4) and the R=2, since R^2 = 4 and R is a positive number.
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Ideas: Compare to the standard equation: (x-h)^2+(y-k)^2 = R^2
2007-08-26 05:31:27 · answer #3 · answered by sahsjing 7 · 2⤊ 0⤋
(x-0)^2(y-4)^2 =2^2.
It represents a circle with centre (0,4) and radius=2.
2007-08-26 05:33:34 · answer #4 · answered by Anonymous · 0⤊ 0⤋
May be written as:-
(x - 0) ² + (y - 4) ² = 2 ²
Centre (0 , 4)
radius 2
2007-08-26 07:21:06 · answer #5 · answered by Como 7 · 0⤊ 0⤋