Question 652851
The equation of the circle in standard form is {{{(x - 5)^2 + (y - 2)^2 = 16}}}.
The center is (5,2). The radius is r = {{{sqrt(16)}}} = 4.
Here's the graph:
{{{drawing(300,300,-2,10,-4,8,grid(1),circle(5,2,4))}}}.
There's no y-intercept. 
To get the x-intercepts, let y = 0 and solve for x.
{{{(x-5)^2 + (0 - 2)^2 = 16}}}
{{{(x - 5)^2 = 12}}}
x - 5 = ±{{{2sqrt(3)}}}
x = 5 ± {{{2sqrt(3)}}}
The coordinates of the x-intercepts are
(5 + {{{2sqrt(3)}}}, 0) and (5 - {{{2sqrt(3)}}}, 0).