# How is the center of a sphere and the point of intersection calculated? How is the point of intersection of the plane y=3 with the sphere (x - 1)^2 + y^2 + (z - 3)^2 <= 9 equal to (1, 3, 3)? Also, how is the center of the sphere x^2 + (y-2)^2 + z^2<=16 which touches the plane z = 4 given by (0, 2, 0)?

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Here the equation of the sphere is (x - 1)^2 + y^2 + (z - 3)^2 <= 9. So a solid sphere has been considered here with the outer surface given by (x - 1)^2 + y^2 + (z - 3)^2=9. Now a plane y=3 intersects this sphere. We see that the plane is tangential to the sphere, this means they meet at just one point.

Also, the radius of the sphere is 3 as 9 is 3^2 or radius^2.

So if the plane y=3 is tangential to the sphere which has a radius 3, the x and y terms should contribute 0. Therefore we have x-1 = 0 and z-3=0 or x=1 and z = 3. This gives the point of intersection as (1,3,3).

Next we have the plane defined by z = 4 and the equation of the solid sphere as x^2 + (y-2)^2 + z^2 <= 16. As the plane is tangential to the sphere they touch at only one point. At this point z = 4. You will also notice that 16 = 4^2, so the radius of the sphere is 4. This implies that the contributions of the other terms involving x and y is zero. So we need to find the center such that x^2 and (y-2)^2 are 0. This is possible if x = 0 and y = 2. Therefore the center is (0,2,0).

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