Frankly, I am concerned that Reflected seems more concerned with learning to use Mathematica than with learning mathetmatics! If you are expected to be able to do problems like this one, you certainly should not need Mathematica to realize that x2+ y2+ z2= 4 is a sphere with center at the origin and radius 2, or that x+ y+ z= 1 is a plane through (0, 0, 1), (0, 1, 0), and (1, 0, 0). Tools are good, crutches are not.
The intersection of the two is a circle with center at (1/3, 1/3, 1/3) (that is the point on x+ y+ z= 1 closest to (0,0,0), the center of the sphere). The distance from (0, 0, 0) to that point is \sqrt{(1/3)^2+ (1/3)^2+ (1/3)^2}= \sqrt{3}/3. From the Pythagorean theorem then, the radius of the circle of intersection is \sqrt{4- 1/3}= \sqrt{33}/3.
It's not really necessary to know all that to find the parameterization of the circle. Here's how I would do it.
From the plane equation, z= 1- x- y. Putting that into the equation of the sphere, x^2+ y^2+ (1- x- y)^2= x^2+ y^2+ 1- 2x- 2y+ x^2+ y^2+ 2xy= 2x^2+ 2y^2+ 2xy- 2x- 2y= 3. Dividing through by 2, x^2+ y^2+ xy- x-y= 3/2. We can eliminate the "xy" term by making the substitutions u= x+ y, v= x- y so that x= (1/2)u+ (1/2)v, y= (1/2)u- (1/2)v and the equation becomes (3/4)u^2- u+ (1/4)v^2= 3, eliminating the "mixed" term. completing the square in "u" part, we have (u- 2/3)^2+ (1/3)v^2= 40/9or (9/40)(u- 2/3)^2+ (3/40)v^2= 1. An obvious parameterization for that is u= (3/2\sqrt{10})cos(\theta)+ 2/3, v= (\sqrt{3}/2\sqrt{10})sin(\theta). Now, work backwards using u= x+ y, v= x- y to get x and y in terms of \theta and then use z= 1- x- y to get z in terms of \theta.<br />
<br />
Modulo any arithmetic errors that should do it.
