Respuesta :
for a circle with center (h,k) and radius, r
the equaiton is
[tex](x-h)^2+(y-k)^2=r^2[/tex]
so
center=(5,2), radius=3
[tex](x-5)^2+(y-2)^2=3^2[/tex]
[tex](x-5)^2+(y-2)^2=9[/tex]
the equaiton is
[tex](x-h)^2+(y-k)^2=r^2[/tex]
so
center=(5,2), radius=3
[tex](x-5)^2+(y-2)^2=3^2[/tex]
[tex](x-5)^2+(y-2)^2=9[/tex]
Equation of the circle;
[tex] (x-a)^{2} + (y-b)^{2} = r^{2} [/tex]
a=5, b=2 and r= 3
[tex] (x-5)^{2} + (y-2)^{2} = 3^{2} [/tex]
[tex] (x-a)^{2} + (y-b)^{2} = r^{2} [/tex]
a=5, b=2 and r= 3
[tex] (x-5)^{2} + (y-2)^{2} = 3^{2} [/tex]