will reward MEDAL for help!

standard form for the equation of a circle?

2x^2+2y^2+8x+7=0

Question

will reward MEDAL for help!


standard form for the equation of a circle?

2x^2+2y^2+8x+7=0

anonymous · Answer

$$2x ^{2}+2y ^{2}+8x+7=0$$

lucaz · Answer

you want the in form (x-a)^2+(y-b)^2=r^2 ?

anonymous · Answer

yes

anonymous · Answer

You have to create perfect squares with the numbers given.
Can you try that?

lucaz · Answer

with these 2 coefficients ?

anonymous · Answer

$$2(x ^{2}+8x+16)+2y ^{2}=35$$

anonymous · Answer

You can observe that if you add +1 to each side of the equation you will have:
$$2x^2+8x+2y^2+7+1=1$$
which is equal to: $$2x^2+8x+2y^2+8=1$$
Common factor is 2 so we have:
$$2(x^2+4x+4+y^2)=1$$
You can observe that $$x^2+4x+4=(x+2)^2$$
Divide both sides by 2 and you will have:
$$(x+2)^2+y^2=1/2$$
and $$1/2=(\sqrt{1/2})^2$$.
Does that help?

anonymous · Answer

ok, how do you find the radius?

anonymous · Answer

The equation is $$(x+2)^2+y^2=r^2$$
where in this case, $$r=\sqrt{1/2}$$
You got it?

anonymous · Answer

yes, thanks

anonymous · Answer

You are welcome! Keep it up! ;)