Parametrizing intersection of a plane and surface I'm working on…

Parametrize the curve which is the intersection of the plane $2x+4y+z=4$ with the surface $z=x^2+y^2$.

I tried eliminating $z$ by plugging it into the first equation and also tried parametrizing each equation before finding the intersection but was not successful. The answer is supposed to be $x=3\cos(t)-1$, $y=3\sin(t)-2$, $z=14-6\cos(t)-12\sin(t)$.

Step 1: As you suggest, eliminate $z$ by plugging it into the first equation:
$$2x + 4y + x^2 + y^2 = 4$$
Now, complete the square. You get the equation of a circle:
$$(x + 1)^2 - 1 + (y + 2)^2 - 4 = 4$$
$$(x + 1)^2 + (y + 2)^2 = 9$$
This is a circle with center $(-1,-2)$ and radius $3$. So $x = -1 + 3\cos(t)$ and $y = -2 + 3\sin(t)$, and $z = 4 - 2x - 4y$.
