Question

can anyone find the center and the radius for
x^2+y^2-6x+2y+9=0

Answer 1

(x-3)^2+(y+1)^2=1
center (3,-1) radius 1

Answer 2

how did you actually work that
or is it a long process im so confused

Answer 3

You need to get it into the standard form (x-h)^2 +(y-k)^2=r^2
I grouped the x-terms and y-terms together individually
x^2-6x +y^2+2y +9=0
it is called completing the square.
x^2-6x since the x^2 coefficient is 1, I take 1/2 of the b coefficient and square it
(1/2)(6)=3. 3^2 = 9
so I need x^2-6x+9 so I used the +9 from the problem
(x^2-6x+9) +(y^2+2y)=0
(x-3)^2 +(y^2+2y ) =0
now I do the same with y^2+2y
(1/2)(2) = 1, 1^2 =1
so I add 1 to both sides of the equation
(x-3)^2 +(y^2+2y +1 ) =0+1
(x-3)^2 + (y+1)^2 = 1 1=1^2
so the center is (3,-1) with r=1

Answer 4

You make it so much easier than my math professor

Answer 5

glad I could help

Answer 6

:)