How do you find the center and radius for $x^2 + y^2 + 8y + 6 = 0$ ?

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Answer 1 · 2016-06-17T10:57:40

cetre at (0,-4) $and radius =$ r = $sqrt10$ .

The General Eqn. of a circle is given by $x^2+y^2+2gx+2fy+c=0,$ provided that $g^2+f^2-c>0.$

Comparing the given eqn. with the general one, we get, $g=0,f=4,c=6,$ so that, the condition $g^2+f^2-c=0+16-6=10>0$ is satisfied.

Then, the centre is $(-g,-f)=(0,-4)$ and radius $=sqrt(g^2+f^2-c)=sqrt10.$

$II^nd$ METHOD

Completing the squares of the given eqn., we have,
$x^2+y^2+8y+16-10-=0.$
$:. (x-0)^2+(y+4)^2=10=(sqrt10)^2.$ ......(1)

Now recall that eqn. of a circle with centre $(h,k)$ & radius $=r$ , is given by,

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

Comparing (1) & (2), we get the cetre at (0,-4) $and$ r = $sqrt10.$

How do you find the center and radius for x^2 + y^2 + 8y + 6 = 0x^2 + y^2 + 8y + 6 = 0?