The radius of a sphere is the distance between the center and any point on the sphere. You know the center, and you know a point on the sphere, so the distance between the two points gives you the radius $r$.
Once you know the radius and the center, the equation of the sphere is very simple. It's almost like the case of a circle on the plane.
To jog your memory: the circle with center $(a,b)$ and radius $r$ has equation
$$(x-a)^2 + (y-b)^2 = r^2.$$
Do you recall now what the equation of a sphere with radius $r$ and center $(a,b,c)$ is?
And since you know the center, and now you know how to find the radius, then the equation of the sphere is...
And as far as moving everything to one side, well, in the case of the circle I gave above, you can just as well write $(x-a)^2 + (y-b)^2 = r^2$ as $(x-a)^2 + (y-b)^2 - r^2 = 0$.
For the second problem, again let's think in terms of the plane. Suppose you want the equation of the circle with center at $(2,10)$ and completely contained in the first quadrant. How would we proceed? Imagine the largest circle: it must just touch at least one of the axes; how far is $(2,10)$ from the axes? It is $2$ away from the $y$-axis, and $10$ away from the $x$-axis. So the largest the radius can be is $2$, otherwise it would "cut through" the $y$-axis. So you'd want the circle with radius $2$ and center in $(2,10)$.
How would you go about doing this for the sphere in the first octant?