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 

where Pt(h,k) is the center and r is the radius

Center of the circle is Pt(-1,3)   (with radius of 2)



 

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