Question

x^2+y^2-2x+6y+9=0 . What is the center of the circle of this equation? (:

Answer 1

x^2+y^2+Dx+Ey+F=0
Centre = ( -D/2 , -E/2)

Answer 2

\[x^2-2x+y^2+6y+9=0\]
Compete the square.
\[(x-1)^2-(1)^2+(y+3)^2-(3)^2+9=0\]
\[(x-1)^2+(y+3)^2-1=0\]
\[(x-1)^2+(y+3)^2=1\]
Centre \(a,b\) and radius \(r\) units is:
\[(x-a)^2+(y-b)^2=r^2\]
soo..centre = 1 and -3

Answer 3

thank you everyone! :D