Question
In the coordinate plane, which equation describes the locus of points 5 units from the point (–2, 2)?
(x + 2)2 + (y – 2)2 = 5
(x – 2)2 + (y + 2)2 = 25
(x – 2)2 + (y + 2)2 = 5
(x + 2)2 + (y – 2)2 = 25
(x + 2)2 + (y – 2)2 = 5
(x – 2)2 + (y + 2)2 = 25
(x – 2)2 + (y + 2)2 = 5
(x + 2)2 + (y – 2)2 = 25
Asked by: USER5759
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247 Answers
Answer (247)
This is a circle centered at (-2,2) with radius 5.
center = (h,k) = (-2,2) so h = -2 and k = 2
radius = r = 5
The equation of any circle is (x-h)^2 + (y-k)^2 = r^2
Plug the three values mentioned into that equation
(x-h)^2 + (y-k)^2 = r^2
(x-(-2))^2 + (y-2)^2 = 5^2
(x+2)^2 + (y-2)^2 = 25
Answer: Choice D
center = (h,k) = (-2,2) so h = -2 and k = 2
radius = r = 5
The equation of any circle is (x-h)^2 + (y-k)^2 = r^2
Plug the three values mentioned into that equation
(x-h)^2 + (y-k)^2 = r^2
(x-(-2))^2 + (y-2)^2 = 5^2
(x+2)^2 + (y-2)^2 = 25
Answer: Choice D
Answer:
This is a circle centered at (-2,2) with radius 5.
center = (h,k) = (-2,2) so h = -2 and k = 2
radius = r = 5
The equation of any circle is (x-h)^2 + (y-k)^2 = r^2
Plug the three values mentioned into that equation
(x-h)^2 + (y-k)^2 = r^2
(x-(-2))^2 + (y-2)^2 = 5^2
(x+2)^2 + (y-2)^2 = 25
Answer: Choice D