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Luba_88 [7]
3 years ago
5

Identify the center of the circle whose equation is (x-2)^2+(y+8)^2=16

Mathematics
2 answers:
levacccp [35]3 years ago
7 0

Answer:

The center of the circle is (2, -8)

Step-by-step explanation:

Given : the equation of circle as (x-2)^2+(y+8)^2=16

We have to find the center of the circle whose equation is  (x-2)^2+(y+8)^2=16

The general equation of a circle is given as (x-h)^2+(y-k)^2=r^2

Where , (h,k) represents the center of the circle and r represents the radius of the circle.

Comparing the given equation of circle  (x-2)^2+(y+8)^2=16 with the general we get,

h = 2 , k = -8 and r = 4

Thus  (x-2)^2+(y+8)^2=16 can be written as (x-2)^2+(y-(-8))^2=4^2

Thus, the center of the circle is (2, -8)

AlekseyPX3 years ago
6 0
The equation of a circle is

(x-h)^{2}+(y-k)^{2} = r^{2}

So this equation you provided has:
 Center: (2,-8)
 Radius: 4
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