Circle graph!! need help with this hard equation >.<?

Question

x^2+2x+y^2-8y+8=0

i dont see how you turn this equation into a cricle >.< please help!

Answer 1

Hi,

x² + 2x + y² - 8y + 8 = 0

Move the constant to the other side.

x² + 2x + y² - 8y = -8

Now complete the square on x and then on y. Multiply ½ times the x term's coefficient. Square the answer. Add that to both sides.

x² + 2x + y² - 8y = -8

½(2) = 1 and 1² = 1, so add 1 to both sides.

x² + 2x + _1_+ y² - 8y = -8 + _1_

½(-8) = -4 and (-4)² = 16, so add 16 to both sides.

x² + 2x + _1_+ y² - 8y + _16_ = -8 + _1_ + _16_

Factor the x terms as a trinomial and then the y terms. Combine terms on the right.

(x + 1)² + (y - 4)² = 9

This is your circle's equation in the form (x - h)² + (y - k)² = r², where (h,k) is the center point of the circle and "r" is the radius.

For your equation (-1,4) is the center and 3 is the radius. Notice the center is always the opposite signs of what you see in the equation.

I hope that helps!! :-)

Answer 2

x^2 + 2x +y^2 - 8y + 8 =0
x^2 + 2x +8= - y^2 +8y
x^2 + 2x + 1 + 7 = -y^2 + 8y +(16-16)
x^2 + 2x + 1+7= -(y^2-8y+16)+16
(x+1)^2+7= 16-(y-4)^2
(x+1)^2+(y-4)^2=16-7=9
(x+1)^2+(y-4)^2= 3^2
this is the eq. of circle, having,
centre- (-1,4)
radius- 3
now I hope you can draw the graph.

Answer 3

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

center(-1,4)
radius=3