Projection of the intersection between a paraboloid and a sphere I am trying to find the projection in the xy-plane from the intersection between a sphere with radius sqrt(3) and the paraboloide z=1/2(x^2+y^2). 

I tried to insert z^2=1/4(x^2+y^2) in the equation for the sphere x^2+y^2+x^2=3, but then I get a equation with x^4 and y^4. From plotting it implicitly, it seems like the answer should be x^2+y^2=2. How can I reduce it to that form?

Thank you.
The paraboloid

$$
z = \frac 12(x^2+y^2)
$$

and the sphere

$$
x^2+y^2+z^2=3
$$

intersect at

$$
z = \frac 12(3-z^2)
$$

or at the feasible $z = 1$ hence the intersection is the circle 

$$
x ^2+y^2=2
$$
Given, 2z=x^2 + y^2

Substitute this in the equation of sphere

Solve the quadratic in z 

You will get z=1, (and imaginary root -3)

This means x-y plane that is shifted towards positive z axis by a unit distance and so the projection on z=1 will be the projection on x-y plane (z=0)

Substitute z=1 in the equation of the sphere.