Given the equation )x+3)^2+y^2=16 , (a) find the center and radius; (b) graph; (c) find the intercepts, if any. Please show all of your work

Question

anonymous · Answer

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At any point $$r^2=x^2+y^2$$
Because of Pythagoras. 

The only thing that would satisfy this equation is a circle.

So, $$r^2=16 \therefore r=4$$

Replacing x with (x+3) just complicates it a little. Just move the centre 3 to the left, negative end. 
So the intercepts are (on the x-axis) -3 (the centre)+4 and -3-4

Which are -7 and 1, which is you insert into the first equation gives y=0