The general equation for the family of circles is
`(x-a)^2 + (y-b)^2 = r^2 `
with `(a,b) ` as center and `r` radius
Let us compare our equation with the general equation
there is only one x term, so a becomes 0, i.e., `(x-a)^2 = (x-0)^2 = x^2`
but there are two terms of y. To find b value we need to find the roots of the quadratic polynomial of `y^2 + 10 y` .
One of the ways to find roots of this polynomial is completing square method.
`(y)^2 + 2*y*5 + (5)^2 -(5)^2` (equating this to `a^2 + 2 ab + b^2` identity)
`= (y+5)^2 -25 `
Now combine `x` and `y` terms and write the equation for the circle
`(x-0)^2 + [y- (-5)]^2 -25 = 0 `
`(x-0)^2 +[y -(-5)]^2 = 5^2 `
`therefore` the center of the circle is (0,-5)
and the radius is 5