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

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Answer 1 · 2017-01-10T00:42:49

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

Given:

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

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:

$x^2-2x+y^2-4/3y-1/3 = 0$

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

$x^2-2x+1+y^2-4/3y+4/9-16/9 = 0$

That is:

$(x-1)^2+(y-2/3)^2 = (4/3)^2$

This is in the form:

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

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

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