Eliminate the parameter of a 
Eliminate the parameter to find a description of the following circles
  or circular arcs in terms of $x$ and $y$. Give the center and radius,
  and indicate the positive orientation.
$x=4\cos{(t)} ,\ y=3\sin{(t)} ;\ 0 \leq t \leq 2\pi$

So,
$\displaystyle x^2=4^2\cos^2{(t)} ,\ y^2=3^2\sin^2{(t)} \implies \frac{x^2}{4^2}=\cos^2{(t)} ,\ \frac{y^2}{3^2}=\sin^2{(t)}$
But I detect no radius. I'm rather confused, on this whole question. It doesn't even explicitly define "the parameter". Insight?
 A: Those equations describe an ellipse, rather than a circle. I'm guessing that by radius they mean length of major and minor axes?
The parameter is $t$, and you're on the right track to eliminate it; your next step is to add the equations 
$$ \frac{x^2}{4^2}=\cos^2{(t)} \\ \frac{y^2}{3^2}=\sin^2{(t)}$$
and use the pythagorean identity.
A: You have not a radius since you are describing an ellipse. Indeed
$$ 1=cos^2(t)+sin^2(t)=\frac{x^2}{16}+\frac{y^2}{9}, $$ then
$$9x^2+16y^2=144.$$
The major semi axis is $4$, and the minor semi axis in $3$.
A: Notice, $$x=4\cos t \implies \cos t=\frac{x}{4}\tag 1$$ & $$y=3\sin t \implies \sin t=\frac{y}{3}\tag 2$$ Now, for eliminating $t$, squaring & adding (1) & (2), we get $$\cos^2t+\sin^2t=\left(\frac{x}{4}\right)^2+\left(\frac{y}{3}\right)^2$$ $$\color{blue}{\frac{x^2}{16}+\frac{y^2}{9}=1}$$ The above equation is in the form of the standard form of equation of an ellipse: $\color{blue}{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$ where, $a=4$ & $b=3$.
Hence, the curve represents an ellipse not a circle, hence we have $$\text{center of ellipse, origin}\equiv(0, 0)$$  $$\text{major axis}, 2a=2\times 4=8$$ $$\text{minor axis}, 2b=2\times 3=6$$
O.P.'s detection is right that there is no radius as that is not a circle but the curve represents an ellipse.  
