equation of ellipse after projection If I have the intersection of $x+z=1$ and $$x^2 +y^2 +z^2=1$$ which is a circle in $O'xyz$. Then I do a projection of this circle on the $O'xy$ plane, it'll be an ellipse. How can I then find the equation of this ellipse?

If $(x,y,z)$ is a point on the circle, then $(x,y,z)$ satisfy both $x+z=1$ and $x^2+y^2+z^2=1$. It's equations are
$$\begin{cases} z=1-x \\ x^2+y^2+(1-x)^2=1\end{cases}$$
Its projection has $z$-coordinate equals $0$ and keep the $x$ and $y$-coordinates. So the equation of the projection on the $xy$-plane is
\begin{align}
x^2+y^2+(1-x)^2&=1\\
2x^2+y^2-2x&=0
\end{align}

Another way to think of the problem.
The centre $C$ of the circle is a point on the plane $x+z=0$. The line joining $C$ and the origin is orthogonal to the plane. It is easy to see that $\displaystyle C=\left(\frac{1}{2},0,\frac{1}{2}\right)$. $C$ is $\displaystyle  \frac{\sqrt{2}}{2}$ unit from the origin. So the radius of the circle is $\displaystyle\sqrt{1-\left(\frac{\sqrt{2}}{2}\right)^2}=\frac{\sqrt{2}}{2}$.
The projection of the circle is an ellipse with centre $\displaystyle \left(\frac{1}{2},0,0\right)$. The major axis of the ellipse is parallel to the $y$-axis, which has the same length as the radius of the circle ($\displaystyle =\frac{\sqrt{2}}{2}$). As the plane makes a $45^\circ$ with the $xy$-plane, the minor axis has length $\displaystyle \frac{\sqrt{2}}{2}\cos45^\circ=\frac{1}{2}$, and it is parallel to the $x$ axis. The equation of the ellipse is 
\begin{align}
\frac{(x-\frac{1}{2})^2}{(\frac{1}{2})^2}+\frac{y^2}{(\frac{\sqrt{2}}{2})^2}&=1\\
(2x-1)^2+2y^2&=1
\end{align}
