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

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Answer 1 · 2017-01-10T00:42:49

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

Given:

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

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

Divide the equation by $- 3$ $-3$ to get:

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

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

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

That is:

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

This is in the form:

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

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

How do you classify the conic -3x^2-3y^2+6x+4y+1=0−3x2−3y2+6x+4y+1=0-3x^2-3y^2+6x+4y+1=0?