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