Consider the points (1,3) and (4,5) in the plane. If we start at the point (1,3) and move 3 units to the right and 2 units up we form a right triangle. This means the distance between my 2 points may be computed using the Pythagorean Theorem. The distance would equal the square root of 3 squared plus 2 squared which is square root of 13. Of course we can do this for any 2 points in the plane and indeed extend this definition to 3 dimensions as well. the distance between points (x1,y1,z1) and (x2,y2,z2) is square root of ((x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2). It is from this same definition that we get the equation of a sphere in 3d. A sphere is the set of all points in 3d which are a fixed distance(called the radius) from another point(called the center). If (a,b,c) is the center of a sphere and r is its radius, then the equation of the sphere is given by (x-a)^2 + (y-b)^2 + (z-c)^2 = r^2.
Now I present a problem to you. How many spheres go through the point (5,4,1) and are tangent to all 3 coordinate planes and what are their equations? The coordinate planes are the xy plane, the yz plane, and the xz plane and to be tangent to all 3 of those planes means the sphere must touch all 3 planes in one and only one point. These points must be in the "middle" of the sphere otherwise they would touch the planes in more than one point. That is to say all 3 of these points must not only be the same distance from the center but also the coordinates of the center must all be the same.
For instance, one point on the sphere is (x,y,0) and from that point we would move r units up to the center. We conclude the center is (r,r,r). The equation of the sphere we're looking for is (x-r)^2 + (y-r)^2 + (z-r)^2 = r^2. The point (5,4,1) must satisfy this equation so (5-r)^2 + (4-r)^2 + (1-r)^2 = r^2. Multiplying these 3 binomials gives us 25 - 10r + r^2 + 16 - 8r + r^2 + 1 - 2r + r^2 = r^2. Combining like terms on the left side simplifies the equation to 3r^2 - 20r + 42 = r^2. Subtracting r^2 from both sides and dividing both sides by 2 leaves us with r^2 - 10r + 21 = 0. If we could factor this we will arrive at an answer. Using my previous lessons on factoring we know we need 2 numbers when multiplied equal 21 and when added equal -10. The numbers we seek are -7 and -3 and we conclude (r-7)(r-3)=0. The 2 solutions are r=7 and r=3 which means there are 2 spheres that satisfy the given conditions. Those spheres have equations (x-3)^2 + (y-3)^2 + (z-3)^2 = 9 and (x-7)^2 + (y-7)^2 + (z-7)^2 = 49.