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36X^2+36y^2-24x-12y=31

Guest Apr 23, 2015

Best Answer 

 #2
avatar+93616 
+10

Do you want to know what shape it is?   Looks like a circle to me.   

 

$$\\36x^2+36y^2-24x-12y=31\\
36x^2-24x+36y^2-12y=31\\
36(x^2-\frac{2x}{3}+y^2-\frac{1y}{3})=31\\
(x^2-\frac{2x}{3})+(y^2-\frac{1y}{3})=\frac{31}{36}\\
(x^2-\frac{2x}{3}+\frac{1}{9})+(y^2-\frac{1y}{3}+\frac{1}{36})=\frac{31}{36}+\frac{1}{9}+\frac{1}{36}\\
(x-\frac{1}{3})^2+(y-\frac{1}{6})^2=1\\$$

$$\\Circle:\qquad centre\;(\frac{1}{3},\frac{1}{6})\qquad radius=1\;unit$$

Melody  Apr 24, 2015
 #1
avatar+20022 
+10

36x^2+36y^2-24x-12y=31

$$\small{\text{$
\begin{array}{rcll}
36x^2+36y^2-24x-12y &=& 31 \\
(36x^2-24x)+(36y^2-12y) &=& 31 \quad & | \quad : 36 \\\\
(x^2-\frac{2}{3}x)+(y^2-\frac{1}{3}y) &=& \frac{31}{36} \\ \\
(x-\frac{2}{6})^2-\frac{4}{36}+(y-\frac{1}{6}y)^2-\frac{1}{36} &=& \frac{31}{36} \\ \\
(x-\frac{1}{3})^2 +(y-\frac{1}{6}y)^2&=& \frac{31}{36} +\frac{4}{36}+\frac{1}{36} \\ \\
(x-\frac{1}{3})^2 +(y-\frac{1}{6}y)^2&=& 1 \\ \\
\end{array}
$}}$$

This is a circle with center $$\small{\text{
$(\frac{1}{3} ,\; \frac{1}{6})
$}}$$
 and the radius = 1

heureka  Apr 24, 2015
 #2
avatar+93616 
+10
Best Answer

Do you want to know what shape it is?   Looks like a circle to me.   

 

$$\\36x^2+36y^2-24x-12y=31\\
36x^2-24x+36y^2-12y=31\\
36(x^2-\frac{2x}{3}+y^2-\frac{1y}{3})=31\\
(x^2-\frac{2x}{3})+(y^2-\frac{1y}{3})=\frac{31}{36}\\
(x^2-\frac{2x}{3}+\frac{1}{9})+(y^2-\frac{1y}{3}+\frac{1}{36})=\frac{31}{36}+\frac{1}{9}+\frac{1}{36}\\
(x-\frac{1}{3})^2+(y-\frac{1}{6})^2=1\\$$

$$\\Circle:\qquad centre\;(\frac{1}{3},\frac{1}{6})\qquad radius=1\;unit$$

Melody  Apr 24, 2015
 #3
avatar+93616 
0

Heureka beat me!     hi Heureka 

Melody  Apr 24, 2015

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