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

Jun 17, 2016

cetre at (0,-4)$\mathmr{and} r a \mathrm{di} u s =$r = $\sqrt{10}$.

#### Explanation:

The General Eqn. of a circle is given by ${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0 ,$ 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 $\left(- g , - f\right) = \left(0 , - 4\right)$ and radius $= \sqrt{{g}^{2} + {f}^{2} - c} = \sqrt{10.}$

$I {I}^{n} d$ METHOD

Completing the squares of the given eqn., we have,
${x}^{2} + {y}^{2} + 8 y + 16 - 10 \equiv 0.$
$\therefore {\left(x - 0\right)}^{2} + {\left(y + 4\right)}^{2} = 10 = {\left(\sqrt{10}\right)}^{2.}$......(1)

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

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2.}$............(2)

Comparing (1) & (2), we get the cetre at (0,-4)$\mathmr{and}$r = $\sqrt{10.}$