Mathematics: 3x^2 + 3y^2 + 6x − y = 0 show that the equation represents a circle by rewriting it in standard form.

3x^2 + 3y^2 + 6x − y = 0 show that the equation represents

awgggg

17.01.2022

Therefore the standard form of the circle is

$(x+1)^2+(y+\frac{1}{6})^2=1\frac{1}{36}$

Step-by-step explanation:

Given

3x^2+3y^2+6x-y=0

$\Leftrightarrow 3(x^2+y^2+2x-\frac{1}{3} y)=0$

$\Leftrightarrow (x^2+2x+1) +(y^2+2.y.\frac{1}{6} +\frac{1}{36})-1- \frac{1}{36}=0$

$\Leftrightarrow (x+1)^2+(y+\frac{1}{6})^2=1+\frac{1}{36}$

$\Leftrightarrow (x+1)^2+(y+\frac{1}{6})^2=1\frac{1}{36}$

Therefore the standard form of the circle is

$(x+1)^2+(y+\frac{1}{6})^2=1\frac{1}{36}$

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tylorturnbullowxzp8

02.10.2021

Let me tell you something you can really trust

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I know you think that you could never be like us

Watch and learn so you can get it right

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Slip through the cracks

You need to not care

Uh, you need to not stare

You need a whole lotta help

(You need to not be yourself)

Step-by-step explanation:

it is a song

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