# How do you find the center and radius for 4x^2+4y^2+36y+5=0?

Feb 22, 2017

Centre: $\left(0 , - \frac{9}{2}\right)$

Radius: $\sqrt{19}$

#### Explanation:

You complete two squares :)

For $4 {x}^{2} + 4 {y}^{2} + 36 y + 5 = 0$, tactically i find it easier to remove the coefficient on the quadratic, so we go at this instead:

$4 \left({x}^{2} + {y}^{2} + 9 y + \frac{5}{4}\right) = 0$

Or:
${x}^{2} + \textcolor{red}{{y}^{2} + 9 y} + \frac{5}{4} = 0$

Completing the square in $y$:
${x}^{2} + \textcolor{red}{\left({\left(y + \frac{9}{2}\right)}^{2} - \frac{81}{4}\right)} + \frac{5}{4} = 0$

$\implies {x}^{2} + {\left(y + \frac{9}{2}\right)}^{2} - 19 = 0$

Or:

${x}^{2} + {\left(y + \frac{9}{2}\right)}^{2} = 19$

$\implies {x}^{2} + {\left(y + \frac{9}{2}\right)}^{2} = {\sqrt{19}}^{2}$