Equation of the projection curve of intersecting spheres How do I find the equation of the projection of the intersection of the spheres $x^{2}+y^{2}+z^{2}=1$ and $ x^{2}+(y-1)^{2}+(z-1)^{2}=1$ onto the $yz$-plane?
All points $(x,y,z)$ on the intersection of the two curves satisfy
$$x^2+y^2+z^2=1\qquad\text{ and }\qquad x^2+(y-1)^2+(z-1)^2=1.$$
In particular, this shows that all such points also satisfy
$$y^2+z^2=(y-1)^2+(z-1)^2,$$
which further simplifies to $z=1-y$. Can you take it from here?
The difference

$$
x^2+y^2+z^2-1-(x^2+(y-1)^2+(z-1)^2-1) = 2(y+z-1) = 0
$$

defines the intersection plane.

Now the intersection plane $y+z-1 = 0$ and the projection plane $O yz$ or $x = 0$ are orthogonal hence, the projection should be a segment. See the picture below when in blue appear the circle from the two spheres intersection and the line as intersection of $Oyz$ and $y+z-1=0$.  This segment limits can be determined thru the intersection of the circles $y^2+z^2=1$ and $(y-1)^2+(z-1)^2=1$