is it (2cos t, 2^0.5 sin t, 2^0.5 sin t)? gives x^2+y^2+z^2=4
Yes, that's one possibility. Since you're taking linear algebra, let's come up with an explanation using matrices and vectors, rather than guessing the parametrization.
You had the right idea initially, where you tried to visualize the plane P = {(x, y, z) | y = z} being half-way in the y-z plane (in red lines in my drawing above).
For simplicity, let's forget about the z-axis and just imagine a circle of radius 2 on the x-y plane (in black lines). This circle can be parametrized by,
[x(t); y(t); z(t)] = [2*sin(t); 2*cos(t); 0], for 0 <= t <= 2*pi.
We see that, if we rotate this x-y plane by 45 degrees about the x-axis, we will get our plane P, and the circle that we want for our problem.
So the question just boils down to what is A, the 3 x 3 matrix that will transform/rotate the x-y plane to the plane P?
Recall that the entries of a matrix come from how a linear map acts upon the basis vectors. From the drawing, we know that we want the following to occur:
A * [2; 0; 0] = [2; 0; 0]
A * [0; 2; 0] = [0; sqrt(2); sqrt(2)]
A * [0; 0; 2] = [0; -sqrt(2); sqrt(2)].
Hence,
A * (2*I) = [2 0 0; 0 sqrt(2) -sqrt(2); 0 sqrt(2) sqrt(2)].
=>
A = 1/2 * [2 0 0; 0 sqrt(2) -sqrt(2); 0 sqrt(2) sqrt(2)].
Now that we know the transformation matrix, we can multiply A and the vector [x(t); y(t); z(t)] above to get the parametrization of the transformed circle.
We get,
A * [x(t); y(t); z(t)] = [2*sin(t); sqrt(2)*cos(t); sqrt(2)*cos(t)], for 0 <= t <= 2*pi.
Note that the transformed y and z values share the same value.
