find the center and radius of the circle whose equation is x^2+y^2-8x+2y+8=0?

Question

find the center and radius of the circle whose equation is x^2+y^2-8x+2y+8=0

and how do you find it?

Answer 1

x^2-8x+16 + y^2+2y+1 = 16+1-8 = 9
(x-4)^2 + (y+1)^2 = 3^2
Center: (4, -1)
Radius: r = 3

Answer 2

x^2 + y^2 – 8x + 2y + 8 = 0
group them:
(x^2 – 8x) + (y^2 + 2y) + 8 = 0

Make the two expressions in the parenthesis
into a perfect square like this form--> (a + b)^2 = a^2 + 2ab + b^2 ,
To do that we add some number on both sides(to preserve equality) that would complete the square
(x^2 – 8x + d ) + (y^2 + 2y + e ) + 8 = 0 + f
(d + e = f--> just numbers)

(x^2 – 8x+16) + (y^2 + 2y+1) + 8 = 0 + 17
(x^2 – 8x+16) + (y^2 + 2y+1) = 17 – 8
(x – 4)^2 + (y + 1)^2 = 9

Equation of Circle: (x - h)^2 + (y - k)^2 = r^2
where r is the radius,
h is distance from origin in the x axis,
k is the distance from the origin in the y axis.
[(h,k) is the center of the circle]

So…
x = 4 , y = -1, r = sqrt(9)

Answer: radius = 3, center (4, -1)

Answer 3

x² + y² - 8x + 2y + 8 = 0
(x² - 8x + 16) + (y² + 2y + 1) = -8 + 16 + 1
(x - 4)² + (y + 1)² = 9 = 3²
center (4,-1) radius 3

Answer 4

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RE:
find the center and radius of the circle whose equation is x^2+y^2-8x+2y+8=0?
find the center and radius of the circle whose equation is x^2+y^2-8x+2y+8=0

and how do you find it?

Answer 5

circle x^2 + y^2 -- 8x + 2y + 8 = 0 is
(x -- 4)^2 + (y + 1)^2 = 9 = 3^2
centre is (4, -- 1) and radius = 3