Well, I have to draw this plane:
$$\frac{1}{4}x^{2}+y^{2}=1$$
How does this look like? Is this just a circle? If so I'm not sure how can I get rid of $\frac{1}{4}$ to make something like $(x-a)^{2}$ out of this.
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That's an ellipse with $,x-$ radius of $;2;$ and $;y-$radius of $;1;$ ... – DonAntonio Aug 03 '13 at 11:11
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2http://en.wikipedia.org/wiki/Ellipse#Canonical_form – lab bhattacharjee Aug 03 '13 at 11:12
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http://math.stackexchange.com/a/316815/3936 – jimjim Aug 03 '13 at 12:05
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As DonAntonio wrote, it is an ellipse. But if you will make a change of coordinates $x'=x/2$ you will get a circle
$$(x')^{2}+y^{2}=1.$$
Boris Novikov
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No, you're equation is not the equation of a circle, but rather that of an ellipse.
A simple modification of the equation $$\frac{1}{4}x^{2}+y^{2}=1$$ gives you the canonical form of an ellipse: $$\dfrac {x^2}{a^2} + \frac{y^2}{b^2} = 1.$$
$$\frac{1}{4}x^{2}+y^{2}=1 \iff \frac {x^2}{4} + \frac{y^2}{1} = 1 \iff \frac{x^2}{(2)^2} + \frac {y^2}{(1)^2} = 1$$
$\qquad\qquad\frac 14 x^2 + y^2 = 1$
$\quad$ 
amWhy
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Edit: This answer makes it easy to see why it is not equation of a circle, the longer less elegant way is below:
To Prove that it is equation of a circle or Not, the equation has to match the form $$(x-a)^2+(y-b)^2-r^2=0$$ now set the equation you have $$\frac14 x^2 + y^2-1=0$$ to the equation of circle and try to find $a,b$ and $r$ :
$$(x-a)^2+(y-b)^2-r^2=\frac14 x^2 + y^2-1$$
Now expand the LHS
$$x^2-2ax+a^2+y^2-2by+b^2-r^2=\frac14 x^2 + y^2-1$$
the resulting system will be inconsistent:
$$a^2+b^2-r^2=-1$$
$$-2ax=0 \implies a = 0$$
$$-2by=0 \implies b = 0$$
so the equation of the circle has to be of the form:
$$x^2+y^2-r^2=0$$ but then this causes the problem:
$$x^2+y^2-r^2=\frac14 x^2 + y^2-1$$ equating coefficient of LHS and RHS we end up with :
for $x$ : $1 = \frac14$ which is absurd, so the equation is not a circle $\square$
