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Verizon [17]
3 years ago
9

3x^2 + 3y^2 + 6x − y = 0

Mathematics
1 answer:
Blizzard [7]3 years ago
7 0

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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