pls help, easy geometry question

x^2 + y^2 + 2x - 6y + 1 = 0
x^2 +2x + 1 + y^2 - 6y = 0
x^2 + 2( x × 1) + 1^2 + y^2 - 2(y×3) + 3^2 -3^2 =0
(x+1)^2 + (y-3)^2 = 3^2
comparing with circle equation
(x-a)^2 + (y - b)^2 = r^2
where (a , b) is centre of circle.
so we get center of circle is
(-1, 3)
Answer:
x²+y²+2x -6y + 1 = 0
x² +2x + 1 + y² - 6y = 0
x² + 2( x × 1) + 1² + y² - 2(y×3) + 3² -3² =0
(x+1)² + (y-3)² = 3²
The circle equation is:
(x-a)² + (y - b)² = r²
(a , b) = centre of circle.
So the centre of the circle is
(-1, 3)
Step-by-step explanation: