We must get it in the standard form:
(x-h)² + (y-k)² = r² where the center is (h,k) and the radius is r.
x² + 10x - 6y + 7 = 0
x² + y² - 2x + 6y - 3 = 0
Get the x terms together and the y terms together
x² - 2x + y² + 6y = 3
We must complete the square on each part, by adding
a number after the x-terms and the y terms. IOW we have
blanks to fill in here:
x² - 2x + __ + y² + 6y + __ = 3 + __ + __
To fill in the first blank on the left and the first blank on the right
side:
1. Multiply the coefficient of x which is -2 by 
getting -1
2. Square this, getting (-1)² = +1
3. Add this to both sides of the equation:
x² - 2x + 1 + y² + 6y + __ = 3 + 1 + __
To fill in the second blank on the left and the second blank on the right
side:
1. Multiply the coefficient of y which is +6 by getting +3
2. Square this, getting (+3)² = +9
3. Add this to both sides of the equation:
x² - 2x + 1 + y² + 6y + 9 = 3 + 1 + 9
The first three terms on the left side of
the equation factor as x²-2x+1 = (x-1)(x-1) = (x-1)²
The last three terms on the left side of
the equation factor as y²+6y+9 = (y+3)(y+3) = (y+3)²
We combine the numbers on the right side
(x-1)² + (y+3)² = 13
Compare that to the standard equation for a circle
(x-h)² + (y-k)² = r² where the center is (h,k) and the radius is r.
We see by comparison that h=1, k=-3 and r = 
So the center is (1,-3) and the radius is .
Edwin
