Question 16395
You need to get your equation into the standard form for a circle whose centre is at (h, k) and whose radius is r.

{{{(x - h)^2 + (y - k)^2 = r^2}}}

You can accomplish this by completing the squares in the x-terms and in the y-terms, then factoring the resulting trinomial in x and factoring the resulting trinomial in y. Start by grouping the x-terms  and grouping the y-terms.
{{{(x^2 - 2x) + (y^2 - 16y) = -1}}} Complete the sqauare in the x-terms by adding the square of half the coefficient of x.  Similarly for the y-terms. Don't forget to add these to both sides of the equation.

{{{(x^2 - 2x + 1) + (y^2 - 16y + 64) = -1 + 1 + 64}}} Simplify and factor the x-trinomial and the y-trinomial.
{{{(x - 1)^2 + (y - 8)^2 = 64}}} Compare this with the standard form:
{{{(x - h)^2 + (y - k)^2 = r^2}}}

The centre of the circle is at (1, 8) and its radius is {{{sqrt(64) = 8}}}