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

Jan 31, 2017

The centre of the circle is $\left(0 , - \frac{9}{2}\right)$ and the radius $= \sqrt{19}$

#### Explanation:

We complete the squares

$4 {x}^{2} + 4 {y}^{2} + 36 y + 5 = 0$

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

$4 {x}^{2} + 4 \left({y}^{2} + 9 y + \frac{81}{4}\right) = - 5 + 81$

$4 {x}^{2} + 4 {\left(y + \frac{9}{2}\right)}^{2} = 76$

Dividing by $4$

${x}^{2} + {\left(y + \frac{9}{2}\right)}^{2} = \frac{76}{4} = 19$

So the centre of the circle is $\left(0 , - \frac{9}{2}\right)$ and the radius $= \sqrt{19}$
graph{4x^2+4y^2+36y+5=0 [-13.72, 14.76, -13.18, 1.07]}