# How do you find the center and radius of the the circle x^2 + y^2 + 6/5x - 8/5y - 8 = 0?

Feb 13, 2016

First factorise the original equation, so that you have it in the form of a circle equation:
${\left(x + \frac{3}{5}\right)}^{2} + {\left(y - \frac{4}{5}\right)}^{2} = 9$
From this we can see that the radius $= \sqrt{9} = 3$, the centre x-coordinate is -3/5=-0.6 and the centre y-coordinate is 4/5=0.8.

#### Explanation:

Starting with the original equation:
${x}^{2} + {y}^{2} + \frac{6}{5} x - \frac{8}{5} y - 8 = 0$
which equals:
${x}^{2} + \frac{6}{5} x + {y}^{2} - \frac{8}{5} y + \left(1 - 9\right)$

= ${x}^{2} + \frac{6}{5} x + \frac{9}{25} + {y}^{2} - \frac{8}{5} y + \frac{16}{25} - 9$

Now factorise the above equation, so that you have it in the form of a circle equation:
${\left(x + \frac{3}{5}\right)}^{2} + {\left(y - \frac{4}{5}\right)}^{2} = 9$

The circle equation is:
${\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} = {r}^{2}$

Where r is the radius. The centre x point and the centre y point are ${x}_{0} \mathmr{and} {y}_{0}$ respectively.

From this we can see that the radius $= \sqrt{9} = 3$, the centre x-coordinate is -3/5=-0.6 and the centre y-coordinate is 4/5=0.8.