How can I make (x+2)^2 +y^2=9 into the standard form of the equation of a circle? I know that I have to subtract 0 from y but that leaves (x+2)^2 x+2 when it is supposed to be x-2. Please help )-:

Question

asnaseer · Answer

use the fact that:$$-(-2)=2$$

asnaseer · Answer

so:$$(x+2)=(x-(?))$$

anonymous · Answer

I did that, I was not sure if it was alright for x to be negative.

asnaseer · Answer

you mean in $(x-a)^2$ you weren't sure if 'a' could be negative?

anonymous · Answer

the equation was $$(x+2)^{2} +y ^{2} = 9$$ I subtracted 0 from y, making the equation $$(x+2)^{2} +(y-0)^{2}=9$$ I then multiplied $$(x+2)^{2}$$ by -1 which equaled $$(-x-2)^{2}$$ because -x is not apart of the formula, I was not sure if that was okay or not

asnaseer · Answer

no - your last step is incorrect. you need to express (x+2) as (x-?)

asnaseer · Answer

so:$$x+2=x-(-?)$$what would ? be for this to be true?

anonymous · Answer

(x-(-2)?

asnaseer · Answer

correct

asnaseer · Answer

so:$$(x+2)^2=(x-(-2))^2$$

anonymous · Answer

okie dokie. I was unaware that I could do that. thank you again. I understand now.

asnaseer · Answer

great - I'm glad I was able to help :)