Question

Please help...I don't know what to do or even where to start.
Given the equation (x+3)^2+y^2=16 (a) find the center and the radius (b) graph (c) find the intercepts, if any.

Answer 1

foil first

Answer 2

(x+3)(x+3)

Answer 3

x=-3 and y=0 center radius =4

Answer 4

the general equation of a circle is:\[(x-a)^2+(y-b)^2=r^2\]where (a,b) is the coordinate of the center of the circle, and r is its radius.

Answer 5

x when x=0 y= +-4

Answer 6

so if you rewrite your equation:\[(x+3)^2+y^2=16\]in the form you get:\[(x-(-3))^2+(y-0)^2=4^2\]from which you can see the center must be at (-3, 0) and the radius must be 4 as @cinar said above.

Answer 7

I am sorry I made a mistake

when x=0 y=+-sqrt7