How to get the equation of an ellipse from a circle that was rotated by a number of degrees? Lets say I have a circle with radius "r" and center "x1,y1" in the X-Y plane. If I rotate the circle by "theta" degrees about the Y axis, I would see an ellipse from the X-Y plane. 
How could I obtain the equation for this ellipse? 
 A: Plane rotated $\theta$ around Y axis:
$$
\mathbb{P} : X x_1 \sin \theta  + Z x_1 \cos \theta = 0
$$
Center and radius must hold:
$$
\xi : (x + x_1 \cos \theta ) ^2  + (y - y_1 ) ^2 + (z - x_1 \sin \theta ) ^2 = r^2
$$
Intersecting both surfaces and Parametrizing  ($\phi$) the curve:
$$
X = (r \sin \phi - x_1 ) \cos \theta \rightarrow Z = - (r\sin\phi - x_1)\sin\theta \quad Y = r\cos \phi + y_1
$$
$$
\frac{(X + x_1 \cos \theta)}{\cos \theta} = r \sin\phi \\
(Y - y_1) = r \cos\phi 
$$
$$
\zeta : \frac{(X + x_1 \cos \theta)^2}{r^2 \cos^2\theta} + \frac{(Y-y_1)^2}{r^2} = 1
$$
A: The original circle has equation $$(x-x_1)^2+(y-y_1)^2=r^2$$
The ellipse with the same centre and with semiaxes $a$ and $b$ has equation $$\frac{(x-x_1)^2}{a^2}+\frac{(y-y_1)^2}{b^2}=1$$
When you rotate the circle by angle $\theta$ about the $y$ axis, the coordinates of the centre become $$(x_1\cos \theta, y_1)$$ and the semiaxes become $r\cos \theta$ and $r$.
So you can write out the equation of the ellipse. I hope that helps.
