Finding the equation of the sphere Find the equation of the sphere which touches the three coordinate planes and the plane 2x+y+2z=6. 
Please help me to solve this problem. What does three coordinate planes mean?
A sphere that touches all the coordinate planes has center of the form $(\pm r, \pm r, \pm r)$. If the sphere lies in the first octant, then we have 
$$\left|\frac{2r+r+2r-6}{3}\right| = r$$ and hence $$4r^2 - 15r + 9 = 0$$ or $r = 3/4$ or $r=3$. Thus the spheres are 
$$(x-3)^2 + (y-3)^2 + (z-3)^2 = 3^2$$
and 
$$(x-3/4)^2 + (y-3/4)^2 + (z-3/4)^2 = (3/4)^2$$
Suppose that the center is $(r, -r, r)$. Then we must have 
$$\left|\frac{2r-r+2r-6}{3}\right| = r$$ 
and hence $(r-2)^2 = r^2$ and $r=1$. Thus the sphere
$$(x-1)^2 + (y+1)^2 + (z-1)^2 = 1$$ 
also satisfies the conditions. Similarly we can discuss the other cases.
I get $7$ solutions in all, considering all octants.
Here is a picture
(it doesn't want to load here, being over 2.7MB).

The centre of your sphere is a point $(\pm r, \pm r, \pm r)$, so that it is a distance $r &gt; 0$ from each of the coordinate planes $x=0$, $y=0$, $z=0$, and
is also at distance $r$ from the plane $2x + y + 2z = 6$.