# How do you classify the conic -3x^2-3y^2+6x+4y+1=0?

Jan 10, 2017

This is a circle with centre $\left(1 , \frac{2}{3}\right)$ and radius $\frac{4}{3}$

#### Explanation:

Given:

$- 3 {x}^{2} - 3 {y}^{2} + 6 x + 4 y + 1 = 0$

Looks like a circle: The coefficients of ${x}^{2}$ and ${y}^{2}$ are the same. Let's rearrange it a little...

Divide the equation by $- 3$ to get:

${x}^{2} - 2 x + {y}^{2} - \frac{4}{3} y - \frac{1}{3} = 0$

Complete the squares for $x$ and $y$...

${x}^{2} - 2 x + 1 + {y}^{2} - \frac{4}{3} y + \frac{4}{9} - \frac{16}{9} = 0$

That is:

${\left(x - 1\right)}^{2} + {\left(y - \frac{2}{3}\right)}^{2} = {\left(\frac{4}{3}\right)}^{2}$

This is in the form:

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

which is the equation of a circle with centre $\left(h , k\right) = \left(1 , \frac{2}{3}\right)$ and radius $r = \frac{4}{3}$