Question 390496
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Hi,        
x^2+2x+y^2-6y+6=0    |completing the squares
(x+1)^2 - 1 + (y-3)^2 -9 + 6 = 0
 (x+1)^2 + (y-3)^2 -4 = 0
 (x+1)^2 + (y-3)^2 = 2^2
Standard Form of an Equation of a Circle is {{{(x-h)^2 + (y-k)^2 = r^2}}}
where Pt(h,k) is the center and r is the radius
Center of the circle is Pt(-1,3)   (with radius of 2)
{{{drawing(300,300, -6, 6, -6, 6, grid(1),
circle(-1, 3,0.2),
circle(-1, 3,2),
graph( 300, 300, -6, 6, -6, 6))}}}