[Will give medal] Find the center and radius of the circle. How do I do this? Equation below:

Question

anonymous · Answer

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

sidsiddhartha · Answer

do u know what is the standard form of a circle?

anonymous · Answer

Yes. $$(x-h)^2+(y-k)^2=r^2$$

sidsiddhartha · Answer

then u can try to complete the square first

anonymous · Answer

4^2=16
$$x^2+8x+16+y^2+$$
Not sure what to do next...

sidsiddhartha · Answer

or what u can do it use anothere orm --
$$x^2+y^2+2fx+2gy+C=0$$
this is also a standard form now compare with this equation

sidsiddhartha · Answer

for this form
center=(-f,-g)
and
$$radius=f^2+g^2-c$$

anonymous · Answer

let's stick with the first one

sidsiddhartha · Answer

so 
u have
$$x^2+y^2+8x+7=0 \and ~the~form ~is--\x^2+y^2+2fx+2gy+c=0 now ~comparing--$$u'll have
$$2f=8 \f=4 \and\ 2g=0\g=0$$
so center is=(-f.-g)=(-4,0)
and radius=$$f^2+g^2-c=4^2+0^2-7=16-7=9$$

anonymous · Answer

you can notice that there is no y term. If we compare our equation with the standard equation.which is this: x^2 + y^2 + 2gx +2fy + c = 0

anonymous · Answer

yeah what @sidsiddhartha said

sidsiddhartha · Answer

if u try to complete the square then--$$x^2+2x*4+4^2+(y-0)^2+7-4^2=0 \(x+4)^2+(y-0)^2=9$$
so center=(-4,0) radius=3

sidsiddhartha · Answer

and sorry i missed some thing for the first process radius for the previous case=$$\sqrt{f^2+g^2-c}$$

sidsiddhartha · Answer

got this?

anonymous · Answer

Yes. Thanks!

sidsiddhartha · Answer

so u can do any one of them :)