Respuesta :
The equation of a circle:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
(h,k) - the coordinates of the center
r - the radius
The center is (4,0), the length of the radius is 2√3.
[tex](x-4)^2+(y-0)^2=(2\sqrt{3})^2 \\ \boxed{(x-4)^2+y^2=12}[/tex]
[tex](x-h)^2+(y-k)^2=r^2[/tex]
(h,k) - the coordinates of the center
r - the radius
The center is (4,0), the length of the radius is 2√3.
[tex](x-4)^2+(y-0)^2=(2\sqrt{3})^2 \\ \boxed{(x-4)^2+y^2=12}[/tex]
The equation of such a circle, described as
set forth in the question, is
(x-4)² + y² = 12 .