I really cannot comment on how the earlier teacher explained things.

Instead, I'll try to explain the concept of a sphere starting from a circle. The equation of a circle drawn in the x-y plane is (x- a)^2 + (y- b)^2 = r^2. This circle has a center (a,b) and the radius is r. It is equivalent to taking a string with length r, anchoring one end at (a,b) and tracing a curve with the other end. This distance of any point on this curve from the center is r. If the same is done in the x-y-z plane or in 3 dimensions, we get the equation of a sphere. Each point on the surface of a sphere has a distance of r from the center.

The use of <= implies that all points within the sphere are also being considered. So (x-a)^2 +(y-b)^2+ (z-c)^2<=r^2 is the equation of a solid sphere. The sphere has a radius r and the center is (a,b,c).

Now when we try to find the point of intersection of a plane with the solid sphere we are looking for points which are present both in the solid sphere as well as on the plane.

If the equation of the plane is x = a, it implies that all points which have an x-coordinate equal to a, form the plane irrespective of what the y and z coordinates are. You have to always remember that all things being discussed here are in 3 dimensions, not just in 2 dimensions as you would with a circle drawn on a sheet of paper.

If a plane is tangential to a sphere, the plane and the sphere touch each other at just a single point. The plane in the problem is y=3 and it is tangential to the sphere (x - 1)^2 + y^2 + (z - 3)^2 <= 9.

So this is a solid sphere. Now the y-coordinate of the point of intersection is y=3 as that is y-coordinate of any point on the plane. You will also notice that 3 is the radius of the sphere. So the point of intersection has the x-coordinate as 1 and the z-coordinate as 3. substituting these values in the equation of the sphere we get (1 - 1)^2 + 3^2 + (3 - 3)^2 = 0+9+0 = 9. Precisely what we want!

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