# How do you find the center and radius of the circle 9x^2 + 54x + 9y^2 − 6y + 55 = 0?

Oct 23, 2016

center $\left(- 3 , \frac{1}{3}\right)$

Radius $\sqrt{3}$

#### Explanation:

Given -

$9 {x}^{2} + 54 x + 9 {y}^{2} - 6 y + 55 = 0$

Let us rewrite the equation in the standard form -

$9 \left({x}^{2} + 6 x\right) + 9 \left({y}^{2} - \frac{2}{3} y\right) = - 55$

$9 \left({x}^{2} + 6 x + 9\right) + 9 \left({y}^{2} - \frac{2}{3} y + \frac{1}{9}\right) = - 55 + 81 + 1$

$9 \left({x}^{2} + 6 x + 9\right) + 9 \left({y}^{2} - \frac{2}{3} y + \frac{1}{9}\right) = 27$

$\left({x}^{2} + 6 x + 9\right) + \left({y}^{2} - \frac{2}{3} y + \frac{1}{9}\right) = \frac{27}{9} = 3$

${\left(x + 3\right)}^{2} + {\left(y - \frac{1}{3}\right)}^{2} = 3$

Now it is in the form of -

(x-h)^2+(y-k)^2=a^2

In this case $\left(h , k\right)$ is the center and $a$ is the radius

Apply this to our case

center $\left(- 3 , \frac{1}{3}\right)$

Radius $\sqrt{3}$