Respuesta :

mathmate
The equation of a circle 
[tex](x-xc)^2+(y-yc)^2=r^2[/tex]
defines a circle centred at (xc,yc) with radius r.

So with (xc,yc)=(0,3), and r=4
the equation becomes 
[tex](x-0)^2+(y-3)^2=4^2[/tex]
[tex]x^2+(y-3)^2=16[/tex]
cheyenneyork

Answer:

b

Step-by-step explanation: